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APPLIED MECHANICS. 



BY 



GAETANO LANZA, S.B., C. & M.E., 

PROFESSOR OF THEORETICAL AND APPLIED MECHANICS, MASSACHUSETTS 
INSTITUTE OF TECHNOLOGY. 



NINTH EDITION, REVISED. 
FIRST THOUSAND. 



NEW YORK: 

JOHN WILEY & SONS, 

London : CHAPMAN & HALL, Limited. 
1905. 






USRARYoi- 


IQMflm&SS 


f#o OoDiess 




OCT. 25 


iyo5 




Gum a. m* m 

COPY S. 



Copyright, 1885, 1900, 1905, 

BY 

GAETANO LANZA. 



ROBEKT DWJMMOND, 1RIN1EK, NEW YORK. 



9" ' 



PREFACE. 



This book is the result of the experience of the writer 
in teaching the subject of Applied Mechanics for the last 
twelve years at the Massachusetts Institute of Technology. 

The immediate object of publishing it is, to enable him to 
dispense with giving to the students a large amount of notes. 
As, however, it is believed that it may be found useful by 
others, the following remarks in regard to its general plan 
are submitted. 

The work is essentially a treatise on strength and stabil- 
ity ; but, inasmuch as it contains some other matter, it was 
thought best to call it "Applied Mechanics/' notwithstanding 
the fact that a number of subjects usually included in trea- 
tises on applied mechanics are omitted. 

It is primarily a text-book ; and hence the writer has endeav- 
ored to present the different subjects in such a way as 
seemed to him best for the progress of the class, even though 
it be at some sacrifice of a logical order of topics. While 
no attempt has been made at originality, it is believed that 
some features of the work are quite different from all pre- 



IV PREFACE. 



vious efforts ; and a few of these cases will be referred to, 
with the reasons for so treating them. 

In the discussion upon the definition of "force," the object 
is, to make plain to the student the modern objections to the 
usual ways of treating the subject, so that he may have a 
clear conception of the modern aspect of the question, rather 
than to support the author's definition, as he is fully aware 
that this, as well as all others that have been given, is open 
to objection. 

In connection with the treatment of statical couples, it 
was thought best to present to the student the actual effect 
of the action of forces on a rigid body, and not to delay this 
subject until dynamics of rigid bodies is treated, as is usually 
done. 

In the common theory of beams, the author has tried to 
make plain the assumptions on which it is based. A little 
more prominence than usual has also been given to the longi- 
tudinal shearing of beams. 

In that part of the book that relates to the experimental 
results on strength and elasticity, the writer has endeavored 
to give the most reliable results, and to emphasize the fact, 
that, to obtain constants suitable for use in practice, we 
must deduce them from tests on full-size pieces. This prin- 
ciple of being careful not to apply experimental results to 
cases very different from those experimented upon, has long 
been recognized in physics, and therefore needs no justifica- 
tion. 

The government reports of tests made at the Watertown 
Arsenal have been extensively quoted from, as it is believed 



PREFACE. 



that they furnish some of our most reliable information on 
these subjects. 

The treatment of the strength of timber will be found to 
be quite different from what is usually given ; but it speaks 
for itself, and will not be commented upon here. 

In the chapter on the " Theory of Elasticity," a combina- 
tion is made of the methods of Rankine and of Grashof. 

In preparing the work, the author has naturally consulted 
the greater part of the usual literature on these subjects ; and, 
whenever he has drawn from other books, he has endeavored to 
acknowledge it. He wishes here to acknowledge the assist- 
ance furnished him by Professor C. H. Peabody of the Massa- 
chusetts Institute of Technology, who has read all the proofs, 
and has aided him materially in other ways in getting out the 
work. 

Gaetano Lanza. 

Massachusetts Institute of Technology, 
April, 1885. 



PREFACE TO THE FOURTH EDITION. 

The principal differences between this and the earlier 
editions consist in the introduction of the results of a large 
amount of the experimental work that has been done during 
the last five years upon the strength of materials. 

The other changes that have been made in the book are not 
a great many, and have been suggested as desirable by the 
author's experience in teaching. 

September, 1890. 



PREFACE TO THE SEVENTH EDITION. 



THE principal improvements in this edition consist in the 
introduction, in Chapter VII, of the results of a considerable 
amount of the experimental work on the strength of materials 
that has been done during the last six years. A few changes 
have also been made in other parts of the book. 

October, 1896. 



PREFACE TO THE EIGHTH EDITION. 



In this edition a considerable number of additional results 
of recent tests, especially upon full-size pieces, have been 
introduced, some of the older ones having been omitted to 
make room for them. 

September, 1900. 



TABLE OF CONTENTS. 



CHAPTER I. 
Composition and Resolution of Forces .......,,.•• i 

CHAPTER II. 
Dynamics . , •••••• 75 

CHAPTER HI. 
Roof-Trusses * * 138 

CHAPTER IV. 
Bfidge-Trusses 184 

CHAPTER V. 
Centre of Gravity .221 

CHAPTER VI. 
Strength of Materials 240 

CHAPTER VII. 

Strength of Materials as Determined by Experimeiti 350 



Vlll TABLE OF CONTENTS. 

CHAPTER VIIL 
Continuous Girders . 743 

CHAPTER IX. 
Equilibrium Curves. — Arches and Domes • 779 

CHAPTER X. 
Theory of Elasticity, and Applications .... ••••«•« 852 



APPLIED MECHANICS. 



CHAPTER I. 
COMPOSITION AND RESOLUTION OF FORCES. 

§ i. Fundamental Conceptions. — The fundamental con- 
ceptions of Mechanics are Force, Matter, Space, Time, and 
Motion. 

§ 2. Relativity of Motion. — The limitations of our natures 
are such that all our quantitative conceptions are relative. 
The truth of this statement may be illustrated, in the case of 
motion, by the fact, that, if we assume the shore as fixed in 
position, a ship sailing on the ocean is in motion, and a ship 
moored in the dock is at rest ; whereas, if we assume the sun 
as our fixed point, both ships are really in motion, as both par- 
take of the motion of the earth. We have, moreover, no means 
of determining whether any given point is absolutely fixed in 
position, nor whether any given direction is an absolutely fixed 
direction. Our only way of determining direction is by means 
of two points assumed as fixed ; and the straight line joining 
them, we are accustomed to assume as fixed in direction. 
Thus, it is very customary to assume the straight line joining 
the sun with any fixed star as a line fixed in direction ; but if 
the whole visible universe were in motion, so as to change the 
absolute direction of this line, we should have no means of 
recognizing it. 



APPLIED MECHANICS. 



§3. Rest and Motion. — In order to define rest and 
motion, we have the following ; viz., — 

When a single point is spoken of as having motion or rest, 
some other point is always expressed or understood, which is 
for the time being considered as a fixed point, and some direc- 
tion is assumed as a fixed direction : and we then say that the 
first-named point is at rest relatively to the fixed point, when 
the straight line joining it with the fixed point changes neither 
in length, nor in direction; whereas it is said to be in motion 
relatively to the fixed point, when this straight line changes in 
length, in direction, or in both. 

If, on the other hand, we had considered the first-named 
point as our fixed point, the same conditions would determine 
whether the second was at rest, or in motion, relatively to the 
first. 

A body is said to be at rest relatively to a given point and 
to a given direction, when all its points are at rest relatively to 
this point and this direction. 

§ 4. Velocity and Acceleration. — When the motion of 
one point relatively to another, or of one body relatively to 
another, is such that it describes equal distances in equal times, 
however small be the parts into which the time is divided, 
the motion is said to be uniform and the velocity constant. 

The velocity, in this case, is the space passed over in a unit 
of time, and is to be found by dividing the space passed over in 
any given time by the time ; thus, if s represent the space 
passed over in time t y and v represent the velocity, we shall 

have 

s 

v = — 
/ 

When the motion is not uniform, if we divide the time into 
small parts, and then divide the space passed over in one of 
these intervals by the time, and then pass to the limit as these 
intervals of time become shorter, we shall obtain the velocity 



FORCE. 3 

Thus, if A* represent the space passed over in the interval of 

time At, then we shall have 

As 
v = limit of — as A/ diminishes, 
A/ 

or 

ds 

In this case the rate of change of velocity per unit of time 
is called the Acceleration, and if we denote it by/, we have 

dv _d 7 s 
f~di~~dt*' 

§ 5. Force. — We shall next attempt to obtain a correct defi- 
nition of force, or at least of what is called force in mechanics. 

It may seem strange that it should be necessary to do this ; 
as it would appear that clear and correct definitions must have 
been necessary in order to make correct deductions, and there- 
fore that there ought to be no dispute whatever over the mean- 
ing of the word force. Nevertheless, it is a fact in mechanics, 
as well as in all those sciences which attempt to deal with the 
facts and laws of nature, that correct definitions are only gradu- 
ally developed, and that, starting with very imperfect and often 
erroneous views of natural laws and phenomena, it is only after 
these errors have been ascertained and corrected by a long 
range of observation and experiment, and an increased range of 
knowledge has been acquired, that exactness and perspicuity 
can be obtained in the definitions. 

Now, this is precisely what has happened in the case of 
force. 

In ancient times rest was supposed to be the natural state 
of bodies ; and it was assumed that, in order to make them 
move, force was necessary, and that even after they had been 
set in motion their own innate inertia or sluggishness would 
cause them to come to rest unless they were constantly urged 



APPLIED MECHANICS. 



on by the application of some force, the bodies coming to rest 
whenever the force ceased acting. 

It was under the influence of these vague notions that such 
terms arose as Force of Inertia, Moment of Inertia, Vis Viva 
or Living Force, etc. 

A number of these terms are still used in mechanics; but 
in all such cases they have been re-defined, such new mean- 
ings having been attached to them as will bring them into 
accord with the more advanced ideas of the present time. 
Such definitions will be given in the course of this work, as 
the necessity may arise for the use of the terms. 

NEWTON'S FIRST LAW OF MOTION. 

Ideas becoming more precise, in course of time there was 
framed Newton's first law of motion ; and this law is as fol- 
lows : — 

A body at rest will remain at rest, and a body in motion will 
continue to move uniformly and in a straight line, unless and 
until some external force acts upon it. 

The assumed truth of this law was based upon the observed 
facts of nature ; viz., — 

When bodies were seen to be at rest, and from rest passed 
into a state of motion, it was always possible to assign some 
cause ; i.e., they had been brought into some new relationship, 
either with the earth, or with some other body: and to this 
cause could be assigned the change of state from rest to motion. 
On the other hand, in the case of bodies in motion, it wa«> seen, 
that, if a body altered its motion from a uniform rectilinear 
motion, there was always some such cause that could be 
assigned. Thus, in the case of a ball thrown from the hand, 
the attraction of the earth and the resistance of the ait soon 
caused it to come to rest. In the case of a ball rolled along 
the ground, friction (i.e., the continual contact and collision with 
the ground) gradually destroyed its motion, and brought it to 



FORCE. 



rest ; whereas, when such resistances were diminished by railing 
it on glass or on the ice, the motion always continued longer : 
hence it was inferred, that, were these resistances entirely 
removed, the motion would continue forever. 

In accordance with these views, the definition of force 
usually given was substantially as follows : — 

Force is that which causes, or tends to cause, a body to change 
its state from rest to motion, from motion to rest, or to change its 
motion as to direction or speed. 

Under these views, uniform rectilinear motion was recog- 
nized as being just as much a condition of equilibrium, or of 
the action of no force or of balanced forces, as rest ; and the 
recognition of this one fact upset many false notions, destroyed 
many incorrect conclusions, and first rendered possible a science 
of mechanics. Along with the above-stated definition of force 
is ordinarily given the following proposition ; viz., — 

Forces are proportional to the velocities that they impart, in a 
unit of time {i.e. to the accelerations that they impart), to the 
same body. The reasoning given is as follows : — 

Suppose a body to be moving uniformly and in a straight 
line, and suppose a force to act upon it for a certain length of 
time t in the direction of the body's motion : the effect of the 
force is to alter the velocity of the body ; and it is only by this 
alteration of velocity that we recognize the action of the force. 
Hence, as long as the alteration continues at the same rate, we 
recognize the same force as acting. 

If, therefore,/ represent the amount of velocity which the 
force would impart in one unit of time, the total increase in 
the velocity of the body will be //; and, if the force now stop 
acting, the body will again move uniformly and in the same 
direction, but with a velocity greater by ft. 

Hence, if we are to measure forces by their effects, it will 
follow that — 

The velocity which a force will impart to a given {or standard) 



APPLIED MECHANICS. 



body in a unit of time is a proper measure of the force. And 
we shall have, that two forces, each of which will impart the 
same velocity to the same body in a unit of time, are equal to 
each other ; and a force which will impart to a given body twice 
the velocity per unit of time that another force will impart to 
the same body, is itself twice as great, or, in other words, — 

Forces are proportional to the velocities that they impart, in a 
unit of time (i.e. to the accelerations that they impart), to the 
same body. 

MODERN CRITICISM OF THE ABOVE. 

The scientists and the metaphysicians of the present time 
are recognizing two other facts not hitherto recognized, and the 
result is a criticism adverse to the above-stated definition of 
force. Other definitions have, in consequence, been proposed ; 
but none are free from objection on logical grounds, and at the 
same time capable of use in mechanics in a quantitative way. 

The two facts referred to are the following ; viz., — 

i°. That all our ideas of space, time, rest, motion, and even 
of direction, are relative. 

2°. That, because two effects are identical, it does not follow 
that the causes producing those effects are identical. 

Hence, in the light of these two facts, it is plain, that, inas- 
much as we can only recognize motion as relative, we can only 
recognize force as acting when at least two bodies are con- 
cerned in the transaction ; and also that if the forces are simply 
the causes of the motion in the ordinary popular sense of the 
word cause, we cannot assume, that, when the effects are equal, 
the causes are in every way identical, although we have, of 
course, a perfect right to say that they are identical so far as 
the production of motion is concerned. 

I shall now proceed, in the light of the above, to deduce a 
definition of force, which, although not free from objection, 
seems as free as any that has been framed. 

It is one of the facts of nature, that, when bodies are by any 



FORCE. 



means brought under certain relations to each other, certain 
tendencies are developed, which, if not interfered with, will 
exhibit themselves in the occurrence of certain definite phe- 
nomena. What these phenomena are, depends upon the nature 
of the bodies concerned, and on the relationships into which 
they are brought. 

As an illustration, we know that if an apple is placed at a 
certain height above the surface of the earth, there is developed 
between the two bodies a tendency to approach each other; 
and if there is no interference with this tendency, it exhihits 
itself in the fall of the apple. If, on the other hand, the apple 
were hung on the hook of a spring balance in the same posi- 
tion as before, the spring would stretch, and there would be 
developed a tendency of the spring to make the apple move 
upwards. This tendency to make the "apple move upwards 
would be just equal to the tendency of the earth and apple to 
approach each other. This would be expressed by saying that 
the pull of the spring is just equal and opposite to the weight 
of the apple. 

As other illustrations of these tendencies developed in 
bodies when placed in certain relations to each other, we have 
the following cases : — 

(a) When two bodies collide. 

(b) When two substances, coming together, form a chemical 
union, as sodium and water. 

(c) When the chemical union is entered into only by raising 
the temperature to some special point. 

Any of these tendencies that are developed by bringing 
about any of these special relationships between bodies might 
properly be called a force ; and the term might properly be, and 
is, used in the same sense in the mental and moral world, as 
well as in the physical. In mechanics, however, we have to 
deal only with the relative motion of bodies ; and hence we 
give the name force only to tendencies to change the relative 



8 APPLIED MECHANICS. 

motion of the bodies concerned ; and this, whether these ten- 
dencies are unresisted, and exhibit themselves in the actual 
occurrence of a change of motion, or whether they are resisted 
by equal and opposite tendencies, and exhibit themselves in 
the production of a tensile, compressive, or other stress in the 
bodies concerned, instead of motion. 

DEFINITION OF FORCE. 

Hence our definition of force, as far as mechanics has to 
deal with it oris capable of dealing with it, is as follows; 
viz., — 

Force is a tendency to change the relative motion of the two 
bodies between which that tendency exists. 

Indeed, when, as in the illustration given a short time ago, 
the apple is hung on the hook of a spring balance, there still 
exists a tendency of the apple and the earth to approach each 
other ; i.e., they are in the act of trying to approach each other ; 
and it is this tendency, or act of trying, that we call the force of 
gravitation. In the case cited, this tendency is balanced by 
an opposite tendency on the part of the spring ; but, were the 
spring not there, the force of gravitation would cause the apple 
to fall. 

Professor Rankine calls force "an action between two bodies, 
either causing or tending to cause change in their relative rest 
or motion ;" and if the act of trying can be called an action, my 
definition is equivalent to his. 

For the benefit of any one who wishes to follow out the 
discussions that have lately taken place, I will enumerate the 
following articles that have been written on the subject : — 

(a) " Recent Advances in Physical Science," by P. G. Tait, 
Lecture XIV. 

(b) Herbert Spencer, " First Principles of Philosophy * 
(certain portions of the book). 



MEASURE OF FORCE. 



(c) Discussion by Messrs. Spencer and Tait, " Nature," Jan. 
2, 9, 16, 1879. 

(d) Force and Energy, "Nature," Nov. 25, Dec. 2, 9, 16, 
1880. 

§6. External Force. — We thus see, that, in order that a 
force may be developed, there must be two bodies concerned 
in the transaction ; and we should speak of the force as that 
developed or existing between the two bodies. 

But we may confine our attention wholly to the motion or 
condition of one of these two bodies ; and we may refer its 
motion either to the other body as a fixed point, or to some 
body different from either ; and then, in speaking of the force, 
we should speak of it as the force acting on the body under 
consideration, and call it an external force.' It is the tendency 
of the other body to change the motion of the body under con- 
sideration relatively to the point considered as fixed. 

§7. Relativity of Force. — In adopting the above-stated 
definition of force, we acknowledge our incapacity to deal with 
it as an absolute quantity ; for we have defined it as a tendency 
to change the relative motion of a pair of bodies. Hence it is 
only through relative motion that we recognize force ; and hence 
force is relative, as well as motion. 

§8. Newton's First Law of Motion. — In the light of 
the above discussion, we might express Newton's first law of 
motion as follows : — 

A body at rest, or in uniform rectilinear motion relatively to 
a given point assumed as fixed, will continue at rest, or in uni- 
form motion in the same direction, unless and until some external 
force acts either on the body in question, or on the fixed point, 
or on the body which furnishes us our fixed direction. This law 
is really superfluous, as it has all been embodied in the defini- 
tion. 

§9. Measure of Force. — We next need some means of 
comparing forces with each other in magnitude; and, subse- 



IO APPLIED MECHANICS. 

quently, we need to select one force as our unit force, by means 
of which to estimate the magnitude of other forces. 

Let us suppose a body moving uniformly and in a straight 
line, relatively to some fixed point ; as long as this motion 
continues, we recognize no unbalanced force acting on it ; 
but, if the motion changes, there must be a tendency to change 
that motion, or, in other words, an unbalanced force is acting 
on the body from the instant when it begins to change its 
motion. 

Suppose a body to be moving uniformly, and a force to be 
applied to it, and to act for a length of time t, and to be so applied 
as not to change the direction of motion of the body, but to 
increase its velocity; the result will be, that the velocity will be 
increased by equal amounts in equal times, and if f represent 
the amount of velocity the force would impart in one unit of 
time, the total increase in velocity will be ft. This results 
merely from the definition of a force ; for if the velocity pro- 
duced in one (a standard) body by a given force is twice as 
great as that produced by another given force, then is the ten- 
dency to produce velocity twice as great in the first case as in 
the second, or, in other words, the first force is twice as great 
as the second. Hence — 

Forces are proportional to the velocities which they will impart 
to a given {or standard) body in a unit of time. 

We may thus, by using one standard body, determine a 
set of equal forces, and also the proportion between different 
forces. 

§10. Measure of Mass. — After having determined, as 
shown, a set of equal (unit) forces, if we apply two of them 
to different bodies, and let them act for the same length of time 
on each, and find that the resulting velocities are unequal, these 
bodies are said to have unequal masses ; whereas, if the result- 
ing velocities are equal, they are said to have equal masses. 

Hence we have the following definitions : — 



RELATION BETWEEN FORCE AND MOMENTUM. II 

1°. Equal forces are those which, by acting for equal times 
on the same or standard body, impart to it equal velocities. 

2°. Equal masses are those masses to which equal forces 
will impart equal velocities in equal times. 

§11. Suppose two bodies of equal mass moving side by- 
side with the same velocity, and uniformly, let us apply to 
one of them a force F in the direction of the body's motion : 
the effect of this force is to increase the velocity with which the 
body moves ; and if we wish, at the same time, to increase 
the velocity of the other, so that they will continue to move 
side by side, it will be necessary to apply an equal force to that 
also. 

We are thus employing a force 2F to impart to the two 
bodies the required increment of velocity. 

If we unite them into one, it still requires a force 2F to 
impart to the one body resulting from their union the re- 
quired increment of velocity : hence, if we double the mass 
to which we wish to impart a certain velocity, we must double 
the force, or, in other words, employ a force which would 
impart to the first mass alone a velocity double that required. 
Hence — 

Forces are proportional to the masses to which they will impart 
the same velocity in the same time. 

§ 12. Momentum. — The product obtained by multiplying 
the number of units of mass in a body by its velocity is called 
the momentum of the body. 

§13. Relation between Force and Momentum. — The 
number of units of momentum imparted to a body in a unit of 
time by a given force, is evidently identical with the number 
of units of velocity that would be imparted by the same force, 
in the same time, to a unit mass. Hence — 

Forces are proportional to the momenta {or velocities per unit 
of mass) which they will generate in a unit of time. 



12 APPLIED MECHANICS. 

Hence, if F represent a force which generates, in a unit of 
time, a velocity/ in a body whose mass is m, we shall have 

F<xmf; 

and, inasmuch as the choice of our units is still under our con- 
trol, we so choose them that 

F = mf; 

i.e., the force F contains as many units' of force as mf contains 
units of momentum ; in other words, — 

The momentum generated in a body in a unit of time by a 
force acting in the direction of the body s motion, is taken as 
a measure of the force. 

§14. Statical Measure of Force. — When the forces are 
prevented from producing motion by being resisted by equal 
and opposite forces, as is the case in that part of mechanics 
known as Statics, they must be measured by a direct comparison 
with other forces. An illustration of this has already been 
given in the case of an apple hung on the hook of a spring 
balance. In that case the pull of the spring is equal in magni- 
tude to the weight of the apple : indeed, it is very customary 
to adopt for forces what is known as the gravity measure, in 
which case we take as our unit the gravitation, or tendency to 
fall, of a given piece of metal, at a given place on the surface 
of the earth ; in other words, its weight at a given place. 

The gravity unit may thus be the kilogram, the pound, or 
the ounce, etc. 

It is evident, moreover, from our definition of force, and the 
subsequent discussion, that whatever we take as our unit of 
mass, the statical measure of a force is proportional to its 
dynamical measure; i.e., the numbers representing the magni- 
tudes of any two forces, in pounds, are proportional to the 
momenta they will impart to any body in a unit of time. 

§15. Gravity Measure of Mass. — If we assume one 
pound as our unit of force, one foot as our unit of length, and 



NEWTON'S SECOND LAW OF MOTION 1 3 

one second as our unit of time, the ratio between the number 
of pounds in any given force and the momentum it will impart 
to a body on which it acts unresisted for a unit of time, will 
depend on our unit of mass ; and, as we are still at liberty to fix 
this as we please, it will be most convenient so to choose it 
that the above-stated ratio shall be unity, so that there shall be 
no difference in the measure of a force, whether it is measured 
statically or dynamically. Now, it is known that a body falling 
freely under the action of its own weight acquires, every second, 
a velocity of about thirty-two feet per second : this number is 
denoted by g, and varies for different distances from the centre 
of the earth, as does also the weight of the body. 

Now, if W represent the weight of the body in pounds, and 
m the number of units of mass in its mass, we must have, in 
order that the statical and dynamical measures may be equal, 

W= mg. 
Hence 

W 

m = —; 

g 

i.e., the number of units of mass in a body is obtained by divid- 
ing the weight in pounds, by the value of g at the place where 
the weight is determined. 

The values of W and of g vary for different positions, but 
the value of m remains always the same for the same body. 

UNIT OF MASS. 

If m = i, then W = g; or, in words, — 

The weight in pounds of the unit of mass (when the gravity 
measure is used) is equal to the value of g in feet per second for 
the same place. 

§ 16. Newton's Second Law of Motion. — Newton's 
second law of motion is as follows : — 



14 APPLIED MECHANICS. 

" Change of momentum is proportional to the impressed mov- 
ing force, and occurs along the straight line in which the force is 
impressed" 

Newton states further in his " Principia : " — 

" If any force generate any momentum, a double force 
will generate a double, a triple force will generate a triple, 
momentum, whether simultaneously and suddenly, or gradually 
and successively impressed. And if the body was moving 
before, this momentum, if in the same direction as the motion, 
is added ; if opposite, is subtracted ; or if in an oblique direc- 
tion, is annexed obliquely, and compounded with it, according 
to the direction and magnitude of the two." 

Part of this law has reference to the proportionality between 
the force and the momentum imparted to the body ; and this 
has been already embodied in our definition of force, and illus- 
trated in the discussion on the measure of forces. 

The other part is properly a law of motion, and may be 
expressed as follows : — 

If a body have two or more velocities imparted to it simulta- 
neously, it will move so as to preserve them all. 

The proof of this law depends merely upon a proper con- 
ception of motion. To illustrate this law when two velocities 
are imparted simultaneously to a body, let us suppose a man 
walking on the deck of a moving ship : he then has two motions 
in relation to the shore, his own and that of the ship. 

Suppose him to walk in the direction of motion of the 
ship at the rate of 10 feet per second, while the ship moves at 
25 feet per second relatively to the shore : then his motion in 
relation to the shore will be 25 + 10 = 35 feet per second. 
If, on the other hand, he is walking in the opposite direction at 
the same rate, his motion relatively to the shore will be 25 — 
10 = 15 feet per second. 

Suppose a body situated at A (Fig. 1) to have two motions 
imparted to it simultaneously, one of which would carry it to B 




POLYGON OF MOTIONS- 1 5 

in one second, and the other to C in one second ; and that it is 
required to find where it will be at the end of one second, and 
what path it will have pursued. 

Imagine the body to move in obedience 
to the first alone, during one second : it 
would thus arrive at B ; then suppose the 
second motion to be imparted to the body, 
instead of the first, it will arrive at the end of the next sec- 
ond at D, where BD is equal and parallel to AC When 
the two motions are imparted simultaneously, instead of suc- 
cessively, the same point D will be reached in one second, 
instead of two; and by dividing AB and AC into the same 
(any) number of equal parts, we can prove that the body will 
always be situated at some point of the -diagonal AD of the 
parallelogram, hence that it moves along AD. Hence follows 
the proposition known as the parallelogram of motions. 

PARALLELOGRAM OF MOTIONS. 

If there be simultaneously impressed on a body two velocities, 
which would separately be represented by the lines AB and AC, 
the actual velocity will be represented by the line AD, which is 
the diagonal of the parallelogram of which AB and AC are the 
adjacent sides. 

§17. Polygon of Motions. — In all the above cases, the 
point reached by the body at the end of a second when the 
two motions take place simultaneously is the same as that which 
would be reached at the end of two seconds if the motions took 
place successively ; and the path described is the straight line 
joining the initial position of the body, with its position at the 
end of one second when the motions are simultaneous. 

The same principle applies whatever be the number of 
velocities that may be imparted to a body simultaneously. 
Thus, if we suppose the several velocities imparted to be 
(Fig. 2) AB, AC, AD, AE, and AF, and it be required to 



1 6 APPLIED MECHANICS. 

determine the resultant velocity, we first let the body move 

A with the velocity AB for one second ; at the 

~-^-*P end of that second it is found at B ; then let 

* c \. it move with the velocity AC only, and at 

/ the end of another second it will be found 

/ at c ; then with AD only, and at the end of 

/ the third second it will be found at d; at the 

yd ' 

/ end of the fourth at e; at the end of the fifth 

~^f at /. Hence the resultant velocity, when all 

FlG - 2 - are imparted simultaneously, is Af, or the 

closing side of the polygon. 

This proposition is known as the polygon of motions. 

POLYGON OF MOTIONS. 

If there be simultaneously impressed on a body any number 
of velocities, the resulting velocity will be represented by the 
closing side of a polygon of which the lines representing tJie 
separate velocities form the other sides. 

§ 1 8. Characteristics of a Force. — A force has three 
characteristics, which, when known, determine it ; viz., Point 
of Application, Direction, and Magnitude. These can be repre- 
sented by a straight line, whose length is made proportional to 
the magnitude of the force, whose direction is that of the 
motion which the force imparts, or tends to impart, and one end 
of which is the point of application of the force ; an arrow-head 
being usually employed to indicate the direction in which the 
force acts. 

§ 19. Parallelogram of Forces. 

Proposition. — If two forces acting simultaneously at the 
same point be represented, in point of application, direction, 
and magnitude, by two adjacent sides of a parallelogram, their 
resultant will be represented by the diagonal of the parallelo- 
gram, drazvn fr'om the point of application of the two forces. 

Proof. — In the last part of § 16 was proved the propo 



PARALLELOGRAM OF FORCES. 



sition known as the Parallelogram of Motions, for the state- 
ment of which the reader is referred to the close of that 
section. 

We have also seen that forces are proportional to the velo- 
cities which they impart, or tend to impart, in a unit of time, 
to the same body. 

Hence the lines representing the two impressed forces are 
coincident in direction with, and proportional to, the lines repre- 
senting the velocities they would impart in a unit of time to 
the same body ; and moreover, since the resultant velocity is 
represented by the diagonal of the parallelogram drawn with 
the component velocities as sides, the resultant force must coin- 
cide in direction with the resultant velocity, and the length of 
the line representing the resultant force will bear to the result- 
ant velocity the same ratio that one of the component forces 
bears to the corresponding velocity. Hence it follows, that the 
resultant force will be represented by the diagonal of the paral 
lelogram having for sides the two component forces. 

§ 20. Parallelogram of Forces : Algebraic Solution. 

Problem. — Given two forces F and F x acting at the same 
point A {Fig. 3), and inclined to each other at an angle 6 ; reqtiired 
the magnitude and direction of the resultant 
force. 

Let AC represent F, AB represent F„ 
and let angle BAC — 0; then will R = AD 
represent in magnitude and direction the 
resultant force. Also let angle DAC = a; then 
angle DAC we have 

AD 2 = AC 2 + CD 2 - 2AC. CDcosACD. 
But 

ACD = 180 - 6 .-. cosACD = -cos* 




'. R 2 = F 2 + F 2 + 2FF, cos 



. R = \F 2 + F 2 + 2FF X cos 0. 



i8 



APPLIED MECHANICS. 



This determines the magnitude of R. To determine its direc- 
tion, let angle CAD = a .\ angle BAD = — a, and we 
shall have from the triangle DAC 



or 



and similarly 



CD : AD = sin C^Z> : sin ACD, 

F t : R = sin a : sin 

F . /, 
sin a = — ^smfl, 



■p 
sin(0 — a) = - sin0. 



EXAMPLES. 



i°. Given F = 47.34, ^ = 7546, = 73 14' 21"; find ^ and a. 
2 . Given ^ = 5.36, F x = 4.27, = 32 10' ; find i? and a. 
3 . Given F = 42.00, F l = 31.00, — 150 ; find j£ and a. 

4 . Given F = 47.00, 7^ = 75.00, = 253 ; find R and a. 



§21. Parallelogram of Forces when 0= 90 . — When 
the two given forces are at right angles to each other, the for- 
mulae become very much simplified, since the parallelogram 
becomes a rectangle. 

From Fig. 4 we at once deduce 




R = >JF* + F t ; 

F, 
sin a = -£, 

F 

COS a = —. 

R 



EXAMPLES. 



i°. Given F = 3.0, F x = 5.0 ; find R and o. 

2°. Given F = 3.0, F x = — 5.0 ; find R and a. 

3 . Given F = 5.0, ^ = 12.0 ; find i? and a. 

4 . Given F = 23.2, F s = 21.3 ; find ^ and a. 



DECOMPOSITION OF FORCES IN ONE PLANE. 1 9 

§ 22. Triangle of Forces. — If three forces be represented, 
hi magnitude and direction, by the three sides of a triangle taken 
in order, then, if these forces be simultaneously applied at one 
point, tJiey will bala?ice each other. 

Conversely, three forces which, when simultaneously applied 
at one point, balance each other, can be correctly represe7ited in 
7nagnitude and direction by the three sides of a triangle taken in 
order. 

These propositions, which rind a very extensive application, 
especially in the determination of the stresses in roof and 
bridge trusses, are proved as follows : — 

If we have two forces, AC and AB (see Fig. 3), acting at the 
point A, their resultant is, as we have already seen, AD ; and 
hence a force equal in magnitude and opposite in direction to 
AD will balance the two forces AC and AB. Now, the sides of 
the triangle ACDA, if taken in order, represent in magnitude 
and direction the force AC, the force CD or AB, and a force 
equal and opposite to AD; and these three forces, if applied at 
the same point, would balance each other. Hence follows the 
proposition. 

Moreover, we have 

AC : CD : DA = sinADC : sin CAD : sin ACD, 
or 

F ' : F t : R = sin (6 — a) : sin a : sin0; 

or each force is, in this case, proportional to the sine of the 
angle between the other two. 

§ 23. Decomposition of Forces in one Plane. — It is 
often convenient to resolve a force into two components, in two 
given directions in a plane containing the force. Thus, suppose 
we have the force R = AD (Fig. 3), and we wish to resolve it 
into two components acting respectively in the directions AC 
and AB ; i.e., we wish to find two forces acting respectively in 
these directions, of which AD shall be the resultant : we 



20 



APPLIED MECHANICS. 



determine these components graphically by drawing a parallelo- 
gram, of which AD shall be the diagonal, and whose sides shall 
have the directions AC and AB respectively. The algebraic 
values of the magnitudes of the compo- 
nents can be determined by solving the 
triangle ADC. In the case when the 
directions of the components are at right 
angles to each other, let the force R 
(Fig. 5), applied at O, make an angle a 
with OX. We may, by drawing the rect- 
angle shown in the figure, decompose R 
into two components, F and F l} along OX and O Y respectively ; 
and we shall readily obtain from the figure, 




Fig. 5. 



F = R cos a, F x = R sin a. 




Fig. 6. 



EXAMPLES. 

i°. The force exerted by the steam upon the piston of a steam-engine 

at the moment when it is in the position shown in the figure is AB = 

iooo lbs. The resistance of the 

guides upon the cross-head DE is 

vertical. Determine the force acting 

along the connecting-rod AC and 

the pressure on the guides ; also 

resolve the force acting along the connecting-rod into 
two components, one along, and the 
other at right angles to, the crank OC. 
2°. A load of 500 lbs. is placed at 
the apex C of the frame ACB: find 

the stresses in AC and CB respectively. 

3 . A load of 4000 lbs. is hung at C, on the crane 

ABC: find the pressure in the boom BC, and the pull 

on the tie AC, where BC makes an angle of 6o° with the horizontal, 

and AC an angle of 15 . 





Fig. 8. 



COMPOSITION OF FORCES IN OXE PLANE. 21 

4°. A force whose magnitude is 7 is resolved into two forces whose 
magnitudes are 5 and 3 : find the angles they make with the given 
force. 

§ 24. Composition of any Number of Forces in One 
Plane, all applied at the Same Point. 

(a) Graphical Solution. — Let the forces be represented 
(Fig. 2) by AB, AC, AD, AE, and AF respectively. Draw Be 
|| and — AC, cd || and = AD, de || and = AE, and ef || and = 
AF; then will Af represent the resultant of the five forces. 
This solution is to be deduced from 
§ 17 in the same way as § 19 is deduced 
from § 16. 

(b) Algebraic Solution. — Let 
the given forces (Fig. 9), of which 
three are represented in the figure, be 
F, F t , F 2 , F v F 4 , etc. ; and let the angles 
made by these forces with the axis OX o*^ x c £ J- 
be a, a n a 2 , a 3 , a 4 , etc., respectively. FlG -9- 
Resolve each of these forces into two components, in the 
directions OX and OY respectively. We shall obtain for the 
components along OX 

OA = Fcosa, OB = ^cosa,, OC = F 2 cosa 2 , etc.; 

and for those along O Y 

OA t = i^sina, OB l = T^sinaj, OC t = F 2 s'ma 2 , etc. 

These forces are equivalent to the following two ; viz., a 
force Fcos a -\- F, cos ttl + F 2 cos a 2 + F 3 cos a 3 + et c along OX, 
and a force i^sin a + F l sin a, -\- F 2 sin a 2 + F 3 sin a 3 + etc. along 
OY. The first may be represented by XFcosa, and the second 
by 'ZFsiria, where 2 stands for algebraic sum. There remains 
only to find the resultant of these two, the magnitude of which 
is given by the equation 

R = ^(^Fcosay + (S^sina) 2 ; 




22 



APPLIED MECHANICS. 



and, if we denote by a^ the angle made by the resultant with 
OX, we shall have 

%F cos a . XF sin a 

COS Of = — , sin Or = 



R 



R 



EXAMPLES. 



r°. Given \ 



F = 47 
^<= 73 
^ 2 = 43 
^3= 23 



a = 21 

a I= 4 8° 
a 2 = 82° 



Find the result- 
ant force and 
its direction. 



Solution. 



F. 


a. 


COS a. 


sin a. 


F COS a. 


Fsma. 


47 
73 
43 
23 


21° 

48° 

82° 

112° 


0.93358 

O.669I3 

O.I39I7 

-O.37461 


0.35837 

0.743I5 
O.99027 

O.92718 


43.87826 

48.84649 

5-9843! 
-8.61603 


16.84339 

54-24995 
42.58161 
21.32414 










90.09303 


134.99909 



. *. Si^cosa = 90.09303, %F sin a = 134.99909, 

.'. R = V(2^cosa) 2 + (2^sina)* _ !6 2 .2976. 

log ^F COS a = I.954691 

\ogR = 2.210331 

log COS Or 
Or 



= 9-744360 

= 56° 17'. 

Observation. — It would be perfectly correct to use the minus sign 
in extracting the square root, or to call R = —162.2976 ; but then we 
should have 

cos or = 90. 93 3 ? an( j s j n ^ _ T 34'999 9 f 
— 162.2976 —162.2976 



or 



ar= 180°+ 5 6°i7' 



= 2 3 6°i/; 



COMPOSITION OF FORCES APPLIED AT SAME POINT. 2$ 

a result which, if plotted, would give the same force as when we call 
R = 162.2976 and 0^ = 56° 17'. 

Hence, since it is immaterial whether we use the plus or the minus sign 
in extracting the square root provided the rest of the computation be 
consistent with it, we shall, for convenience, use always plus. 



2°. 


^=4, 


a = 77°, 




F> = 3, 


a, = 82°, 




F 2 = 10, 


<*2= 1 63°, 




«- 5, 


a 3 = 2 75°- 


s°. 


*•= 5, 


a = COS -I £j 




F*= 4, 


a t = O, 


» 


F*= 3, 


02 = 90°. 



§ 25. Polygon of Forces. — If arty number of forces be 
represented in magnitude and direction by the sides of a polygon 
taken in order, then, if these forces be simultaneously applied at 
one point, they will balance each other. 

Conversely, any number of forces which, when simultaneously 
applied at one point, balance each other, can be correctly repre- 
sented in magnitude and direction by the sides of a polygon taken 
in order. 

These propositions are to be deduced from § 24 {a) in the 
same way as the triangle of forces is deduced from the parallelo- 
gram of forces. 

§ 26. Composition of Forces all applied at the Same 
Point, and not confined to One Plane. — This problem can 
be solved by the polygon of forces, since there is nothing in 
the demonstration of that proposition that limits us to a plane 
rather than to a gauche polygon. 

The following method, however, enables us to determine 
algebraic values for the magnitude of the resultant and for its 
direction. 



24 



APPLIED MECHANICS. 




Fig. io. 



We first assume a system of three rectangular axes, OX, 

OY, and 0Z (Fig. io), whose origin 
is at the common point of the given 
forces. Now, let OE — F be one 
of the given forces. First resolve 
it into two forces, OC and OD, the 
first of which lies in the z axis, and 
the second perpendicular to 0Z> 
or, as it is usually called, in the z 
plane ; the plane perpendicular to 
OX being the x plane, and that 
perpendicular to OY the y plane. 
Then resolve OD into two com- 
ponents, OA along OX, and OB along OY. We thus obtain 
three forces, OA, OB, and OC respectively, which are equivalent 
to the single force OE. These three components are the edges 
of a rectangular parallelopiped, of which OE = Eis the diagonal 
Let, now, 

angle EOX = a, EOY = (3, and EOZ = y ; 

and we have, from the right-angled triangles EOA, EOB, and 
EOC respectively, 

OA = i^cosa, OB = Ecos/3, OC = i^cosy. 
Moreover, 

OA 2 + OB 2 = OD 2 and OD 2 + OC 2 = OE 2 
.-. OA 2 + OB 2 + OC 2 = OE 2 , 

and by substituting the values of OA, OB, and OC, given above, 

we obtain 

cos 2 a + cos 2 j3 + cos 2 y = i ; 

a purely geometrical relation existing between the three angles 
that any line makes with three rectangular co-ordinate axes. 

When two of the angles a, p, and y are given, the third can 
be determined from the above equation. 



COMPOSITION OF FORCES APPLIED AT SAME POINT. 25 



Resolve, in the same way, each of the given forces into 
three components, along OX, OY, and OZ respectively, and we 
shall thus reduce our entire system 
of forces to the following three 
forces : — 



i°. A single force 2F cos a along OX. 
2 . A single force 'SFcosfi along OY. 
3 . A single force ^F cos y along OZ. 

We next proceed to find a sin- 
gle resultant for these three forces. 
Let (Fig. 11) 

OA = X^cosa 
OB = ZFcosfi, 
OC = XFcosy. 




Fig. i] 



Compounding OA and OB, we find OD to be their resultant ; 
and this, compounded with OC, gives OE as the resultant of 
the entire system. Moreover, 



or 



OE* = OD 2 4- OC 2 = OA 2 + OB 2 + OC 2 , 
R* = (XFcosu) 2 + (%Fcos(3) 2 + (%Fcosy)* 
R == V(2^cosa)*+ (S^cos/?) 2 + (XF cosy) 2 ; 



and if we let EOX =z a r , EOY = J3 r , and EOZ — y r , we shall 
have 

OA XFcosa 2FcosB , XFcosy 
C ° S ar = 0£ = — ^ — ' ° 0S ^ r = ~R' ° OSyr = ~R' 

This gives us the magnitude and direction of the resultant. 

The same observation applies to the sign of the radical for 
R as in the case of forces confined to one plane. 



26 



APPLIED MECHANICS. 



DETERMINATION OF THE THIRD ANGLE FOR ANY ONE FORCE. 

When two of the angles a, /?, and y are given, the cosine of- 
the third may be determined from the equation, — 

cos 2 a + cos 2 /? •+- cos 2 y = i ; 

but, as we may use either the plus or the minus sign in extract* 
ing the square root, we have no means of knowing which of 
the two supplementary angles whose cosine has been deduced 
is to be used. 

Thus, suppose a = 45 , j3 = 6o°, then 



cosy = ±Vi — i 

,\ y = 6o°, Or I20 C 



i= ±* 



but which of the two to use we have no means of deciding. 

This indetermination will be more clearly seen from the fol- 
lowing geometrical considerations : — 

The angle a (Fig. 12), being given as 45 °, locates the line 

representing the force on a right 
circular cone, whose axis is OX, 
and whose semi-vertical angle is 
AOX=BOX=4S°. On the other 
hand, the statement that /? = 6o° 
locates the force on another right 
circular cone, having OY for axis, 
and a semi-vertical angle of 6o°; 
both cones, of course, having their 
vertices at O. Hence, when a and 
/? are given, we know that the line 
representing the force is an element of both cones ; and this is 
all that is given. 

(a) Now, if the sum of the two given angles is less than 
90 , the cones will not intersect, and the data are consequently 
inconsistent. 




Fig. 12. 



DETERMINATION OF THE THIRD ANGLE. 



27 



(b) If, on the other hand, one of the given angles being 
greater than 90 , their difference is greater than 90 , the cones 
will not intersect, and the data are again inconsistent. 

(c) If a + (3 = 90 , the cones are tangent to each other, 
and 7 == 90 . 

(d) If a + /? > 90 , and a — /3ory3 — «< 90 , the cones 
intersect, and have two elements in common ; and we have no 
means of determining, without more data, which intersection 
is intended, this being the indetermination that arises in the 
algebraic solution. 



EXAMPLES. 

\F =63 a= 53 /3 = 4 2 c 

I. Given -J F x = 49 a = 87 y = 72* 

[ F 2 = 2 = 70 y = 45 ( 



Find the magnitude 
and direction of 
the resultant. 



Solution. 



p 


a. 


0- 


y« 


COS a. 


COS/3. 


COSy. 


F COS a. 


F cos /3. 


F COS y. 


63 

49 
2 


53° 
87° 


42° 

70 


72 
45° 


O.60182 
O.O5234 
0.61888 


O.743I4 
O.94961 
O.34202 


O.29250 
O.30902 
O.707 1 1 


37.91466 
2.56466 
I.23776 


46.81782 

46.53089 

0.68404 


18.42750 
I5.I4198 
I.41422 
















4I.7I708 

S^cos a 


94-03275 
2^cos 3 


34-98370 

IF cosy 



R = V / (XFcos.a) 2 + (XFcos/?) 2 + (XF cosy) 2 = 108.6569. 
log S^cosa = 1.620314 log 2Fcos3 — 1.973279 log 2/^cos y = 1.543866 
log j? = 2.036057 log R = 2.036057 log ^ = 2.036057 



log cos a r =9.584257 log cos 3 r =9.937222 log cos y r =9.507809 
a r = 67 25' 20" 3 r = 30° 4' 14" jy = 71 13' 5" 



28 



APPLIED MECHANICS. 



2. 



F. 


a. 


(3. 




/?. 


a. 


P- 


4-3 


47° 2' 


65° 7' 


3- 


5 


9 o° 


9°° 


87-5 


88° 3' 


io° 5' 




7 


0° 




6.4 


68° 4 ' 


83° 2' 




4 
75 


73° 


o° 



45' 



§ 27. Conditions of Equilibrium for Forces applied at a 
Single Point. 

i°. When the forces are not confined to one plane, we have 
already found, for the square of the resultant, 

R* = (XFcosa) 2 4- (ZFcos/3) 2 4- (S^cosy) 2 . 

But this expression can reduce to zero only when we have 

XFcosa = o, 2,Fcos(3 — o, and XFcosy = o; 

for the three terms, being squares, are all positive quantities,, 
and hence their sum can reduce to zero only when they are 
separately equal to zero. 

Hence : If a set of balanced forces applied at a single point 
be resolved into components along three directions at right angles 
to each other, the algebraic stun of the components of the forces 
along each of the three directions must be equal to zero, and con- 
versely. 

2°. When the forces are all confined to one plane, let that 
plane be the z plane ; then y = 90 in each case, and 

.-. /3 = 90 - a 

.*. cos P = sin a 

/. R* = (S^cosa) 2 + (XFsina)*. 

Hence, for equilibrium we must have 

(XF cos a) 2 4- (XFsina) 2 = o; 



STATICS OF RIGID BODIES. 29 

and, since this is the sum of two squares, 

Si^cosa = o, and *%F sin a = o. 

Hence : If a set of balanced fojxes, all situated in one plane, 
a?id acting at one point, be resolved into components along two 
directions at right angles to each other, and in their own plane, 
the algebraic sum of the components along each of the tzvo given 
directions must be equal to zero respectively; and conversely. 

§ 28. Statics of Rigid Bodies. — A rigid body is one that 
does not undergo any alteration of shape when subjected to 
the action of external forces. Strictly speaking, no body is 
absolutely rigid ; but different bodies possess a greater or less 
degree of rigidity according to the material of which they are 
composed, and to other circumstances. When a force is ap- 
plied to a rigid body, we may have as the result, not merely a 
rectilinear motion in the direction of the force, but, as will be 
shown later, this may be combined with a rotary motion ; in 
short, the criterion by which we determine the ensuing motion 
is, that the effect of the force will distribute itself through the 
body in such a way as not to interfere with its rigidity. 

What this mode of distribution is, we shall discuss here- 
after ; but we shall first proceed to some propositions which can 
be proved independently of this consideration. 

§ 29. Principle of Rectilinear Transferrence of Force in 
Rigid Bodies. — If a force be applied to a rigid body at the 
point A (Fig. 13) in the direction AB, 
whatever be the motion that this force 
would produce, it will be prevented from 
taking place if an equal and opposite 
force be applied at A, B, C, or D, or at FlG - I3 - 

any point along the line of action of the force : hence we have 
the principle that — 

The point of application of a force acting on a rigid body, 
may be transferred to any other point which lies in the fate of 




30 



APPLIED MECHANICS. 



action of the force, and also in the body, without altering the 
resulting motion of the body, although it does alter its state of 
stress. 

§ 30. Composition of two Forces in a Plane acting at 
Different Points of a Rigid Body, and not Parallel to Each 
Other. — Suppose the force F (Fig. 14) to be applied at A, and 
Fj at B, both in the plane of the paper, and acting on the rigid 
body abcdef Produce the lines of direction of the forces till 
they meet at O, and suppose both F and F l to act at O. Con- 
struct the parallelogram ODHE, where OD = F and OF = F, ; 

then will OH = R rep^ 
resent the resultant 
force in magnitude and 
in direction. Its point 
of application may be 
conceived at any point 
along the line OH, as 
at C, or any other 
point ; and a force 
equal and opposite to 
OH, applied at any point of the line OH, will balance F at A, 
and F y at B. 

The above reasoning has assumed the points A, B, C and 
O, all within the body : but, since we have shown, that when 
this is the case, a force equal and opposite to R at C will bal- 
ance F2X A, and F l at B, it follows, that were these three forces 
applied, equilibrium would still subsist if we were to remove 
the part bafeghc of the rigid body ; or, in other words, — 

The same construction holds even when the point O falls out- 
side the rigid body. 

§31. Moment of a Force with Respect to an Axis Per- 
pendicular to the Force. 

Definition. — The moment of a force with respect to an 
axis perpendicular to the force, and not intersecting it, is the 




Fig. 14. 



EQUILIBRIUM OF THREE PARALLEL FORCES. 



31 




product of the force by the common perpendicular to (shortest 
distance between) the force and the axis. 

Thus, in Fig. 15 the moment of F about 
an axis through O and perpendicular to the 
plane of the paper is F(OA). The sign of 
the moment will depend on the sign attached 
to the force and that attached to the perpen- 
dicular. These will be assumed in this book Fig. 15. 
in such a manner as to render the following true ; viz., — 

The moment of a force with respect to a,7i axis is called posi- 
tive when, if the axis were supposed fixed, the force would cause 
the body on which it acts to rotate around the axis in the direc- 
tion of the hands of a watch as 
seen by the observer looking at 
the face. It will be called nega- 
tive when the rotation would take 
place in the opposite direction. 

§ 32. Equilibrium of Three 
Parallel Forces applied at 
Different Points of a Rigid 
Body. — Let it be required to 
find a force (Fig. 16) that will 
balance the two forces F at A, 
and F t at B. Apply at A and B 
respectively, and in the line AB, 
the equal and opposite forces Aa 
and Bb. Their introduction will 
produce no alteration in the 
body's motion. 

The resultant of F and Aa 

is Af that of F, and Bb is Bg. 

Compound these by the method 

of § 30, and we obtain as result- 

A force equal in magnitude and opposite in direction 




Fig. 16. 



ant ce. 



32 



APPLIED MECHANICS. 



to ce, applied at any point of the line eC, will be the force 
required to balance Fat A and F t at B; and, as is evident from 
the construction, this line is in the plane of the two forces. 
Moreover, by drawing triangle/^/ equal to Bbg, we can readily 
prove that triangles oce and Afl are equal : hence the angle oce 
equals the angle fAl, and R is parallel to F and F„ Also 
R = ce = ch + he = AK + iS7 = F + X 



and 












Bb Bb' 



.-. since ^4# = i?#, 

AC _F t F_ F\ _ F+ F> . F__F±__ F + F \ 

~BC~"F " BC AC AB " Cr Cq qr 

where qr is any line passing through C. 

Hence we have the following propositions ; viz., — 
If three parallel forces balance each other, — 
i°. They must lie in one plane. 

2°. The middle one must be equal in magnitude and opposite 
F in direction to the sum of the other 

two. 

3°. Each force is proportional to the 
distance between the lines of direction 

of the other two as measured on any 

line intersecting all of them. 

The third of the above-stated con- 
ditions may be otherwise expressed, 

thus : — 

The algebraic sum of the moments 

of the three forces about any axis perpendicular to the forces 

must be zero. 



j_ . 



RESULTANT OF A PAIR OF PARALLEL FORCES. 33 

Proof. — Let F, F lt and R (Fig. 17) be the forces ; and let 
the axis referred to pass through O. Draw OA perpendicular 
to the forces. Then we have 

F(OA) + F t (OB) = F(OC + CA) + F v (OC - BC) 

= (F + F s ) OC + F(AC) - F t (BC). 

But, from what we have already seen, 

F + F t = -R 

and 

BC AC 
.'. F(AC) = FABC) 
.-. F(OA) + F t (OB) = -R(OC) + o 
.-. ^((9^) + F v {OB) +R(OC) = o, 

or the algebraic sum of the moments of the forces about the 
axis through is equal to zero. 

§ 33. Resultant of a Pair of Parallel Forces. — In the 
preceding case, the resultant of any two of the three forces 
F, F lt and R, in Fig. 16 or Fig. 17, is equal and opposite to the 
third force. Hence follow the two propositions : — 

I. If two parallel forces act in the same direction, their 
resultant lies in the plane of the forces, is equal to their sum, 
acts in the same direction, and cuts the line joining their points 
of application, or any common perpendicular to the two forces, 
at a point which divides it internally into two segments in- 
versely as the forces. 

II. If two unequal parallel forces act in opposite directions, 
their resultant lies in the plane of the forces, is equal to their 
difference, acts in the direction of the larger force, and cuts the 
line joining their points of application, or any common perpen- 
dicular to them, at a point which (lying nearer the larger force) 



34 APPLIED MECHANICS. 

divides it externally into two segments which are inversely as 
the forces. 

Another mode of stating the above is as follows : — 

i°. The resultant of a pair of parallel forces lies in the plane 
of the forces. 

2°. It is equal in magnitude to their algebraic sum, and coin- 
cides in direction with the larger force. 

3°. The moment of the resultant about an axis perpendicu- 
lar to the plane of the forces is equal to the algebraic sum of 
the moments about the same axis. 

EXAMPLES. 

i. Find the length of each arm of a balance such that i ounce at 
the end of the long arm shall balance i pound at the end of the short 
arm, the length of beam being 2 feet, and the balance being so propor- 
tioned as to hang horizontally when unloaded. 

2. Given beam =28 inches, 3 ounces to balance 15. 

3. Given beam = 36 inches, 5 ounces to balance 25 ounces. 



MODE OF DETERMINING THE RESULTANT OF A PAIR OF PARALLEL 
FORCES REFERRED TO A SYSTEM OF THREE RECTANGULAR 
AXES. 

Let both forces (Fig. 18) be parallel to OZ ; then we have, 
from what has preceded, 

L = ii = F±*i = * where R = F + F x . 
be ab ae ac 

But from the figure 

be ab F ' F, 



or 



*3 


~"" •A'2 *^2 Osf \ CX"i «^2 t^"2 *^I 




Fx 2 — Fxi = F 1 x l — FiX 2 




.\ (F+F^x^ Fx t + F x x v 




Fx 2 = Fx l -f F,x $ ; 






RESULTANT OF NUMBER OF PARALLEL FORCES. 



35- 



and similarly we may prove that 



or 



i°. The resultant of two parallel forces is parallel to the 
forces and equal to their algebraic sum. 




Fig. 18. 



2°. The moment of the resultant with respect to OX is 
equal to the algebraic sum of their moments with respect to 
OX; and likewise when the moments are taken with respect 
to OK 

§ 34. Resultant of any Number of Parallel Forces. — 
Let it be required to find the resultant of any number of paral- 
lel forces. 

In any such case, we might begin by compounding two of 
them, and then compounding the resultant of these two with a 
third, this new resultant with a fourth, and so on. Hence, for 
the magnitude of any one of these resultants, we simply add 
to the preceding resultant another one of the forces ; and for 
the moment about any axis perpendicular to the forces, we add 



36 



APPLIED MECHANICS. 



to the moment of the preceding resultant the moment of the 
new force. 

Hence we have the following facts in regard to the resultant 
of the entire system : — 

I °. The resultant will be parallel to the forces and equal to 
their algebraic sum. 

2°. The moment of the resultant abotit any axis perpendicular 
to the forces will be equal to the algebi-aic sum of the moments 
of the forces about the same axis. 

The above principles enable us to determine the resultant 
in all cases, except when the algebraic sum of the forces is 
equal to zero. This case will be considered later. 

§ 35. Composition of any System of Parallel Forces 
y when all are in One Plane. — 
Refer the forces to a pair of rect- 
angular axes, OX, OY (Fig. 19), 
and assume OY parallel to the 
forces. 

The forces and the co-ordinates 
of their lines of direction being as 
indicated in the figure, if we denote 
by R the resultant, and by x the 
co-ordinate of its line of direction, 
we shall have, from the preceding, 

R = %F; (1) 

and if moments be taken about an 
axis through O, and perpendicular 



F 5 


• 


\ 


F« 

1 


;, F. 




K ar 5 


x 4 


F4 


c a 


X 


Xl 



Fig. 



to the plane of the forces, we shall also have 

Rx Q = IFx. 
Hence 

R = ^F and x Q = 



(2) 






determine the resultant in magnitude and in line of action, 
except when %F = o, which case will be considered later. 



EQUILIBRIUM OF ANY SET OF PARALLEL FORCES. tf 
* 

§ 36, Composition of any System of Parallel Forces not 
confined to One Plane. — Refer the forces to a set of rect- 
angular axes so chosen that OZ is parallel to their direction. 
If we denote the forces by F lt F 2i F v F 4 , etc., and the co-ordinates 
of their lines of direction by (x„ j/,), (x 2 , y 2 ), etc., and if we 
denote their resultant by R, and the co-ordinates of its line of 
direction by (x Q , y ), we shall have, in accordance with what has 
been proved in § 34, — 

1°. The magnittide of the resultant is equal to the algebraic 
sum of the forces, or 

R = 2F. 

2°. The moment of the resultant about OY is equal to the 
mm of the moments of the forces about OY, or 



x, 



£F = %Fx. 



3 . The moment of the resultant about OX is equal to the 
gitm of the moments about OX, or 

yoZF = $Fy. 

Hence 

r, ^j 77 XFx ^IFy 

R=%F, ^=-, *=jj£ 

determine the resultant in all cases, except when %F = o. 

§ 37. Conditions of Equilibrium of any Set of Parallel 
Fcrces. — If the axes be assumed as before, so that OZ is 
parallel to the forces, we must have 

2F = o, %Fx = o, and %Fy = o. 

To prove this, compound all but one of the forces. Then equilib- 
rium will subsist only when the resultant thus obtained is equal 
and directly opposed to the remaining force ; i.e, it must be 
equal, and act along the same line and in the opposite direction. 
Hence, calling R a the resultant above referred to, and (x a , y a ) 
the co-ordinates of its line of direction, and calling F n the 



$8 APPLIED MECHANICS. 

remaining force, and (x w y n ) the co-ordinates of its line of direc* 
tion, we must have 

Ra = —F n , X a = X n , y a = JK«, 

. • . R a + F n = o, R a x a + F n x n = o, R a y a + F n y n = o, 
.-. %F = o, XFx = o, %Fy = o. 

When the forces are all in one plane, the conditions become 

2F = o, %Fx == o. 

§ 38. Centre of a System of Parallel Forces. — The 

resultant of the two parallel forces F and F t (Fig. 20), ap- 
plied at A and B respectively, is a force R = F -f- F n whose 
line of action cuts the line AB at a point C, 
which divides it into two segments inversely as 
the forces. If the forces F and F s are turned 
through the same angle, and assume the posi- 
tions AO and BO x respectively, the line of 
action of the resultant will still pass through 
fig. 20. £ which is called the centre of the two parallel 

forces F and F y . Inasmuch as a similar reasoning will apply 
in the case of any number of parallel forces, we may give the 
following definition : — 

The centre of a system of parallel forces is the point through 
which the line of action of the resultant always passes, 710 matter 
how the forces are turned, provided only — 

i°. Their points of application remain the same. 
2°. Their relative magnitudes are unchanged. 
3 . They remain parallel to eacJi other. 

Hence, in finding the centre of a set of parallel forces, we 
may suppose the forces turned through any angle whatever, and 
the centre of the set is the point through which the line of 
action of the resultant always passes. 




DISTRIBUTED FORCES. 



39 



§ 39. Co-ordinates of the Centre of a Set of Parallel 
Forces. — Let F 1 (Fig. 21) be one of the forces, and (x lt jr lt z x ) 

the co-ordinates of its point 



of application. Let F 2 be 
another, and (x 2 , y 2 , z 2 ) co- 
ordinates of its point of 
application. Turn all the 
forces around till they are 
parallel to OZ, and find the 
line of direction of the re- 
sultant force when they are 
in this position. The co- 
ordinates of this line are 



x = 



%Fx 
2F' 



2 

''■■ 7f 

qj^ 

Oj X J- 

y, / 



Fig. 2i. 



Jo = 



_ S^V 



^F' 



and, since the centre of the system is a point on this line, the 
above are two of the co-ordinates of the centre. Then turn 
the forces parallel to OX, and determine the line of action of 
the resultant. We shall have for its co-ordinates 



y ° ~ 2F' ° ~ %F* 



Hence, for the co-ordinates of the centre of the system, we 
have 



2F' 



Jo = 



_ ^5 

2F' 



%F' 



When %F = o the co-ordinates would be 00, therefore such 
a system has no centre. 

§ 40. Distributed Forces While we have thus far as- 
sumed our forces as acting at single points, no force really acts 
at a single point, but all are distributed over a certain surface 



40 APPLIED MECHANICS. 

or through a certain volume ; nevertheless, the propositions 
already proved are all applicable to the resultants of these 
distributed forces. We shall proceed to discuss distributed 
forces only when all the elements of the distributed force are 
parallel to each other. As a very important example of such a 
distributed force, we may mention the force of gravity which 
is distributed through the mass of the body on which it acts. 
Thus, the weight of a body is the resultant of the weights of 
the separate parts or particles of which it is composed. As 
another example we have the following : if a straight rod be 
subjected to a direct pull in the direction of its length, and if 
it be conceived to be divided into two parts by a plane cross- 
section, the stress acting at this section is distributed over the 
surface of the section. 

§41. Intensity of a Distributed Force. — Whenever we 
have a force uniformly distributed over a certain area, we obtain 
its intensity by dividing its total amount by the area over which 
it acts, thus obtaining the amount per unit of area. 

If the force be not uniformly distributed, or if the intensity 
vary at different points, we must adopt the following means 
for finding its intensity. Assume a small area containing the 
point under consideration, and divide the total amount of force 
that acts on this small area by the area, thus obtaining the 
mean intensity over this small area : this will be an approxima- 
tion to the intensity at the given point ; and the intensity is the 
limit of the ratio obtained by making the division, as the area 
used becomes smaller and smaller. 

Thus, also, the intensity, at a given point, of a force which 
is distributed through a certain volume, is- the limit of the 
ratio of the force acting on a small volume containing the 
given point, to the volume, as the latter becomes smaller and 
smaller. 

§42. Resultant of a Distributed Force. — i°. Let the 
force be distributed over the straight line AB (Fig. 22), and 



RESULTANT OF A DISTRIBUTED FORCE. 




let its intensity at the point E where AE = x, be represented 
by EF ' = p -— <f>(x), a function of x ; 
then will the force acting on the por- 
tion Ee = Ax of the line be pAx : and 
if we denote by R the magnitude of 
the resultant of the force acting on the 
entire line AB, and by x Q the distance 
of its point of application from A, we shall have 

R = %pAx approximately, 
or 

R = fpdx exactly ; 

and, by taking moments about an axis through A perpendicular 
to the plane of the force, we shall have 

XoR = ^x(pAx) approximately, 
or 

x Q R = fpxdx exactly ; 

whence we have the equations 

fpxdx 
R = fpdx, *o = J -JJ^- 

2°. Let the force be distributed over a plane area EFGEt 

(Fig. 23), let this area be re- 
ferred to a pair of rectangular 
axes OX and OY y in its own 
plane, and let the intensity 
of the force per unit of area 
at the point P, whose co- 
ordinates are x and y, be 
/ = 4>(x, y) ; then will pAxAy 
be approximately the force act- 
ing on the small rectangular 
area AxAy. Then, if we rep- 
resent by R the magnitude of 
the resultant of the distributed force, and by x Q , y Q , the co-ordi- 




DC 
Fig. 23. 



4 2 



APPLIED MECHANICS. 



nates of the point at which the line of action of the resultant 
cuts the plane of EFGH, we shall have 

R = 2/AjcAj approximately, 

Xo R = %x(p±x±y) " 

yo R = %ytpHxby) " 

or, as exact equations, we shall have 

R = ffpdxdy, 
Jjpxdxdy = Sfpydxdy 

Xo = J7pdxW' SSpdxdy' 

*° Let the force be distributed through a volume let this 
volume be referred to a system of rectangular axes, OX, OY, 
and OZ let A V represent the elementary volume, whose co- 
linatt are ,, * I and let p = *. * *) be the intensity o 
the force per unit of volume at the point (*, y, z) then, if we 
represent by R the magnitude of the resultant, and by x , y„z„ 
the co-ordinates of the centre of the distributed force we shall 
have from the principles explained in § 38 and § 3* the approx- 
imate equations 

and these give, on passing to the limit, the exact equations 

fpxdV Spyiv _ SpdV_ 

R = fpJV, Xo = J -j^y> y° = JpW' Z ° SpdV 

8 4 , Centre of Gravity. - The weight of a body, or system 
of bodies, is the resultant of the weight of the separate parts 
or particles into which it may be conceived to be divided , and 
the .centre of gravity of the body, or system o bod es, is the 
centre of the above-stated system of parallel forces, ie the 
point through which the resultant always passes, no mattertiow 
The forces are turned. The weight of any one particle is the 
force which gravity exerts on that particle: hence, if we repre- 



FORCE APPLIED TO CENTRE OF STRAIGHT ROD. 43 

sent the weight per unit of volume of a body, whether it be 
the same for all parts or not, by w, we shall have, as an 
approximation, 

W = 2wA V, x = , y Q = \ , z Q = ; 

and as exact equations, 

fwxdV fwydV fwzdV 

w = fwdVi Xo = 7 ^ v , >'° = 7^F' 0o== /wF' (I) 

where f'F denotes the entire weight of the body, and x of y Q , z oy 
the co-ordinates of its centre of gravity. 

If, on the other hand, we let M == entire mass of the body, 
dM = mass of volume dV, and m = mass of unit of volume, 
we shall have 

W — Mg, w = mg, wdV = mgdV = gdM. 

Hence the above equations reduce to 

n , rM fxdM fydM fzdM . . 

Equations (1) and (2) are both suitable for determining the 
centre of gravity; one of the sets being sometimes most con- 
venient, and sometimes the other. 

§ 44. Centre of Gravity of Homogeneous Bodies If 

the body whose centre of gravity we are seeking is homogeneous, 
or of the same weight per unit of volume throughout, we shall 
have, that w = a constant in equations (1) ; and hence these 
reduce to 

w=wfdv x-l^l y -Jy*K % .-<!*L 

§45. Effect of a Single Force applied at the Centre of 
a Straight Rod of Uniform Section and Material. — If a 

straight rod of uniform section and material have imparted to it 




44 APPLIED MECHANICS. 

2l motion, such that the velocity imparted in a unit of time to 
each particle of the rod is the same, and if we represent this 
velocity by/, then if at each point of the rod, we lay off a line 
xy (Fig. 24) in the direction of the motion, 
and representing the velocity imparted to that 
point, the line bounding the other ends of 
the lines xy will be straight, and parallel to the 
rod. If we conceive the rod to be divided 
into any number of small equal parts, and 
denote the mass of one of these parts by AM, then will f\M 
contain as many units of momentum as there are units of force 
in the force required to impart to this particle the velocity 
f in a unit of time ; and hence fAM is the measure of this 
force. 

Hence the resultant of the forces which impart the velocity 
/ to every particle of the rod will have for its measure 

fM, 

where M is the entire mass of the rod ; and its point of applica- 
tion will evidently be at the middle of the rod. 

It therefore follows that — 

The effect of a single force applied at the middle of a straight 
rod of uniform section and material is to impart to the rod a 
motion of translation in the directio7i of the force, all points of 
the rod acquiring eqtial velocities ill equal times. 

§ 46. Translation and Rotation combined. — Suppose that 
we have a straight rod AB (Fig. 25), and suppose that such a 
force or such forces are applied to it as will impart to the point 
A in a unit of time the velocity Aa, and to the point B the 
(different) velocity Bb in a unit of time, both being perpendicu- 
lar to the length of the rod. It is required to determine the 
motion of any other point of the rod and that of the entire 
rod. 



TRANSLATION AND ROTATION COMBINED. 



45 




Fig. 25. 



Lay off Aa and Bb (Fig. 25), and draw the line ab, and pro- 
duce it till it meets AB produced 
in 0: then, when these velocities 
Aa and Bb are imparted to the 
points A and B, the rod is in the 
act of rotating around an axis 
through perpendicular to the plane of the paper ; for when a 
body is rotating around an axis, the linear velocity of any point 
of the body is perpendicular to the line joining the point in 
question with the axis (i.e., the perpendicular dropped from the 
point in question upon the axis), and proportional to the dis- 
tance of the point from the axis. 

Hence : If the velocities of two of the points in the rod are 
given, and if these are perpendicular to the rod, the motion 
of the rod is fixed, and consists of a rotation about some axis 
at right angles to the rod. 

Another way of considering this motion is as follows : Sup- 
pose, as before, the velocities of the points A and B to be 

represented by Aa and Bb respec- 
tively, and hence the velocity of 
any other point, as x (Fig. 26), to 
be represented by xy, or the length 
of the line drawn perpendicular to 
AB, and limited by AB and ab. 
Then, if we lay off Aa t — Bb, = \{Aa + Bb) = Cc, and draw 
ajb xy and if we also lay off Aa 2 = a t a, and Bb 2 = bjb, we shall 




Fig. 26. 



have the following relations ; viz., — 



Aa 



Aa x — Aa 2 



Bb = Bb, + Bb 2 , 
xy = xy, — xy 2 , 



etc., 



or we may say that the actual motion imparted to the rod in a 
unit of time may be considered to consist of the following two 

parts : — 



46 APPLIED MECHANICS: 

i°. A velocity of translation represented by Aa l} the mean 
velocity of the rod ; all points moving with this velocity. 

2°. A varying velocity, different for every different point, 
and such that its amount is proportional to its distance from 
C, the centre of the rod, as graphically shown in the triangles 
Aa 2 CBb 2 . In other words, the rod has imparted to it two 
motions : — 

i°. A translation with the mean velocity of the rod. 

2°. A rotation of the rod about its centre. 

§47. Effect of a Force applied to a Straight Rod of 
Uniform Section and Material, not at its Centre. — If the 
force be not at right angles to the rod, resolve it into two com- 
ponents, one acting along the rod, and the other at right angles 
to it. The first component evidently produces merely a trans- 
lation of the rod in the direction of its length : hence the second 
component is the only one whose effect we need to study. 

To do this we shall proceed to show, that, when such a rod 
has imparted to it the motion described in § 46, the single re- 
sultant force which is required to impart 
this motion in a unit of time is a force 
acting at right angles to the rod, at a point 
different from its centre ; and we shall de- 
fig. 27. termine the relation between the force and 

the motion imparted, so that one may be deduced from the 
other. 

Let -A be the origin (Fig. 27), and let 

Ac = x y cd ' = dx. 

AB — I — length of the rod. 

ce =f= velocity imparted per unit of time at distance x 
from A. 

Aa = f„Bl>=f 2 . 

w ==., weight per unit of length. 

m ~ mass per unit of length = — . 




EFFECT OF FORCE APPLIED TO A STRA/GHT ROD. 47 

W — entire weight of rod. 

M = entire mass of rod = — . 

g 
R = single resultant force acting for a unit of time to 

produce the motion. 
x Q — distance from A to point of application of R. 

Then we shall have, 

Hence, from § 42, 

] fdx = **(area AabB) = -(/, +/a)/ = ~(/x +/,). (1) 

o 2 2 

C l m M 

Xo R = mXfxdx = -(/ + 2 / a )/» = y(/ + 2/ 2 )/. (2) 

I /, + 2/ a , 



3 /i +/a 



(3) 



We thus have a force i?, perpendicular to AB, whose mag- 
nitude is given by equation (1), and whose point of application 
is given by equation (3) ; the respective velocities imparted by 
the force being shown graphically in Fig. 27. 



EXAMPLES. 

I. Given Weight of rod = W ' = 100 lbs., 
Length of rod = 3 feet, 

Assume £ = 32 feet per second, 

Force applied = R = 5 lbs., 
Point of application to be 2.5 feet from one end; 

determine the motion imparted to the rod by the action of the force for 
one second. 



4$ APPLIED MECHANICS. 

Solution. 
Equation (i) gives us, 



g /ioo\/J-/ 



3 
Equation (2) gives, 

(2.5) (5) = (2S) g) ( 3 )(/ + 2/ 2 ), or/ r + 2 / 2 = 8 

.-. / 2 = 4.8, / = -1.6. 

Hence the rod at the end nearest the force acquires a velocity of 4.8 
feet per second, and at the other end a velocity of —1.6 feet per 
second. The mean velocity is, therefore, 1.6 feet per second; and we 
may consider the rod as having a' motion of translation in the direc- 
tion of the force with a velocity of 1.6 feet per second, and a rotation 
about its centre with such a speed that the extreme end (i.e., a point 
I feet from the centre) moves at a velocity 4.8 — 1.6 = 3.2 feet per 

second. Hence angular velocity = — = 2.14 per second = i22°.6 

per second. 

2. Given £F= 50 lbs., /= 5 feet. It is desired to impart to it, 
in one second, a velocity of translation at right angles to its length, of 5 
feet per second, together with a rotation of 4 turns per second : find the 
force required, and its point of application. 

3. Assume in example 2 that the velocity of translation is in a 
direction inclined 45 ° to the length of the rod, instead of 90 . Solve, 
the problem. 

4. Given a force of 3 lbs. acting for one-half a second at a distance 
of 4 feet from one end of the rod, and inclined at 30 to the rod : 
determine its motion. 

5. Given the same conditions as in example 4, and also a force 
of 4 lbs., parallel and opposite in direction to the 3-lb. force, and acting 
also for one-half a second, and applied at 3 feet from the other end : 
determine the resulting motion. 



MOMENT OF THE FORCES CAUSING ROTATION. 49 

6. Given two equal and opposite parallel forces, each acting at right 
angles to the length of the rod, and each equal to 4 lbs., one being 
applied at 1 foot from one end, and the other at the middle of the rod ; 
find the motion imparted to the rod through the joint action of these 
forces for one-third of a second. 

§ 48. Moment of the Forces causing Rotation. — Re- 
ferring again to Fig. 26, and considering the motion of the 
rod as a combination of translation and rotation, if we take 
moments about the centre C, and compare the total moment 
of the forces causing the rotation alone, whose accelerations 
are represented by the triangles aa x cb x b, with the total moment 
of the actual forces acting, whose accelerations are represented 
by the trapezoid AabB, we shall find these moments equal to 
each other ; for, as far as the forces represented by the rectangle 
are concerned, every elementary force m(xy x )dx on one side of 
the centre C has its moment {Cx)\m(xy x )dx\ equal and opposite 
to that of the elementary force at the same distance on the other 
side of C: hence the total moment of the forces represented 
graphically by the rectangle Aa x b x B is zero, and hence — 

The moment about C of those represented by the trapezoid 
equals the moment of those represented by the triangles. 

Hence, from the preceding, and from what has been pre- 
viously proved, we may draw the following conclusions : — 

i°. If a force be applied at the centre of the rod, it will 
impart the same velocity to each particle. 

2°. If a force be applied at a point different from the centre, 
and act at right angles to its length, it will cause a translation 
of the rod, together with a rotation about the centre of the rod. 

3 . In this latter case, the moment of the forces imparting 
the rotation alone is equal to the moment of the single resultant 
force about the centre of the rod, and the velocity of translation 
imparted in a unit of time is equal to the number of units of 
force in the force, divided by the entire mass of the rod. 



50 



APPLIED MECHANICS. 



§ 49. Effect of a Pair of Equal and Opposite Parallel 
Forces applied to a Straight Rod of Uniform Section and 
Material. — Suppose the rod to be AB (Fig. 28), and let the 
two equal and opposite parallel forces be Dd and Ee, each equal 
to F, applied at D and E respectively. 
The mean velocity imparted in a unit 

of time by either force will be — ; and, 

from what we have already seen, the trap- 
ezoid AabB will furnish us the means of 
representing the actual velocity imparted 
to any point of the rod by the force Dd. 
The relative magnitudes of Aa and Bb, the 
accelerations at the ends, will depend, of 
course, on the position of D ; but we shall 




always have 



Cc = \(Aa + Bb) = -| 



Fig. 28. 



quantity depending only on the magnitude 

of the force. So, likewise, the trapezoid Aa x b x B will represent 

the velocities imparted by the force Ee ; and while the relative 

magnitude of Aa x and Bb x will depend upon the position of E, 

27 
we shall always have Cc, = \{Aa x -f- Bb,) — — . Hence, since 

Cc = Cc u the centre C of the rod has no motion imparted to it 
by the given pair of forces, hence the motion of the rod is one 
of rotation about its centre C 

The resulting velocity of any point of the rod will be the 
difference between the velocities imparted by the two forces ; 
and if these be laid off to scale, we shall have the second 
figure. Hence — 

A pair of equal and opposite parallel forces, applied to a 
straight rod of uniform section and material, produce a rota- 
tion of the rod about its centre. Also, — 

Such a rotation about the centre of the rod cannot be pro- 



EFFECT OF STATICAL COUPLE ON STRAIGHT ROD. 5 I 

duced by a single force, but requires a pair of equal and op- 
posite parallel forces. 

§ 50. Statical Couple. — A pair of equal and opposite 
parallel forces is called a statical couple. 

§51. Effect of a Single Force applied at the Centre of 
Gravity of a Straight Rod of Non-Uniform Section and 
Material. — In the case of a straight rod of non-uniform sec- 
tion and material, we may consider the rod as composed of a 
set of particles of unequal mass : and if we imagine each par- 
ticle to have imparted to it the same velocity in a unit of time, 
then, using the same method of graphical representation as 
before (Fig. 24), the line ab, bounding the other ends of the 
lines representing velocities, will be parallel to AB ; but if we 
were to represent by the lines xy, not the velocities imparted, 
but the forces per unit of length, the line bounding the other 
ends of these forces would not, in this case, be parallel to AB. 
Moreover, since these forces are proportional to the masses, and 
hence to the weights of the several particles, their resultant 
would act at the centre of gravity of the rod. Hence — 

A force applied at the centre of gravity of a straiglit rod will 
impart the same velocity to each point of the rod ; i.e. } will im- 
part to it a motion of translation only. 

§ 52. Effect of a Statical Couple on a Straight Rod of 
Non-Uniform Section and Material. — Let such a rod have 
imparted to it only a motion of rotation about its centre of 
gravity, and let us adopt the same modes of graphical repre- 
sentation as before. 

Let the origin be taken at O (Fig. 29), 
the centre of gravity of the rod. 

Let Aa = f = velocity imparted to A. 
Bb = f 2 — velocity imparted to B. 
OA = a, OB = b, OC = x. 
CD = / = velocity imparted to C. 
dM — elementary mass at C. FlG . 2Q# 




52 APPLIED MECHANICS. 

Then, from similar triangles, we have 

/ = / -lx=jx ) 
a b 

and hence for the force acting on dM we have 

dF = (CB)dx =f±xdM. 
a 

Hence the whole force acting on AO, and represented graph- 
ically by Aa.O, is 

a, 



• /» x ■--- a 

- / xdM, 

f *Jx = o 



and that acting on OB, and represented by BOb lt is 

4 2 {*xdM = f- f X xdM. 
bjx = -b ajx = -i 



Hence for the resultant, or the algebraic sum, of the two, we 
have 






xdM. 
b 



But from § 43 we have for the co-ordinate x Q of the centre of 
gravity of the rod 



x 



*x = a 

xdM 

-b 
x Q = 



M ' 
but, since the origin is at the centre of gravity, we have 

x = o, 
and hence 

f*x = a 

.'. R = o. 



JOx = a 
xdM = o 
x=-b 



Hence the two forces represented by Aa,0 and Bb,0 are equal 
in magnitude and opposite in direction : hence the rotation 
about the centre of gravity is produced by a Statical Couple. 



MEASURE OF THE ROTATORY EFFECT 53 

Now, a train of reasoning similar to that adopted in the case 
of a rod of uniform section and material will show that a single 
force applied at some point which is not the centre of gravity 
of the rod will produce a motion which consists of two parts ; 
viz., a motion of translation, where all points of the rod have 
equal velocities, and a motion of rotation around the centre of 
gravity of the rod. 

§53. Moment of a Couple. — The moment of a statical 
couple is the product of either force by the perpendicular dis- 
tance between the two forces, this perpendicular distance being 
called the arm of the couple. 

§54. Measure of the Rotatory Effect. — Before proceed- 
ing to examine the effect of a statical couple upon any rigid 
body whatever, we will seek a means of measuring its effect in 
the cases already considered. 

The measure adopted is the moment of the couple ; and, in 
order to show that it is proper to adopt this measure, it will be 
necessary to show — - 

That the moment of the couple is proportional to the angu- 
lar velocity imparted to the same rod in a unit of time ; and 
from this it will follow — 

That two couples in the same plane with equal moments will 
balance each other if one is right-handed and the other left-handed 

If we assume the origin of co-ordinates at C (Fig. 30), the 
centre of gravity of the rod, and "if we 
denote by a the angular velocity imparted 
in a unit of time by the forces F and — F, 
and let CD = x lt CE = x 2 , then we have 
for the linear velocity of a particle situated 
at a distance x from C the value 



The force which will impart this velocity in a unit of time to 

the mass dM is 

axdM. 




54 APPLIED MECHANICS. 

The total resultant force is 

afxdM, 

which, as we have seen, is equal to zero. The moment of the 
elementary force about C is 

x{axdM) = ax 2 dM, 
and the sum of the moments for the whole rod is 

afx 2 dM, 

and this, as is evident if we take moments about C, is equal H 

Fxi — Fx 2 = F(x, — x 2 ) = F{DE). 

Now, fx 2 dM is a constant for the same rod : hence any quan- 
tity proportional to F(DE) is also proportional to a. 

The above proves the proposition. 

Moreover, we have 

F(DE) = afx*dM 
_ F(DE) 
fx*dM' . 

whence it follows, that when the moment of the couple is given, 
and also the rod, we can find the angular velocity imparted in 
a unit of time by dividing the former by fx 2 dM. 

§ 55. Effect of a Couple on a Straight Rod when the 
Forces are inclined to the Rod. — We shall next show that 
the effect of such a couple is the same as that of a couple of 
equal moment whose forces are perpen- 
dicular to the rod. 

In this case let AD and BC be the 
forces (Fig. 31). The. moment of this 
couple is the product of AD by the per- 
pendicular distance between AD and BC, 

Fig. 31. r . 

the graphical representation of this being 
the area of the parallelogram ADBC. 




EFFECT OF A STATICAL COUPLE ON A RIGID BODY. 55 

Resolve the two forces into components along and at right 
angles to the rod. The former have no effect upon the motion 
of the rod : the latter are the only ones that have any effect 
upon its motion. The moment of the couple which they form 
is the product of Ad by AB, graphically represented by paral- 
lelogram AdBb ; and we can readily show that 

ADBC = AdBb. 

Hence follows the proposition. 

§ 56. Effect of a Statical Couple on any Rigid Body. — 

Refer the body (Fig. 32) to three rectan- 
gular axes, OX, OY, and OZ, assuming 
the origin at the centre of gravity of the 
body, and OZ as the axis about which 
the body is rotating. Let the mass of the 
particle P be AM, and its co-ordinates be 
x, y, z. 

Then will the force that would impart fig. 32. 

to the mass AM the angular velocity a in a unit of time be 

arAM, 
where r = perpendicular from P on OZ, or 




r = \Jx 2 -f y 2 . 

This force may be resolved into two, one parallel to O Y ana 
the other to OX; the first component being axAM, and the. 
second ay AM. 

Proceeding in the same way with each particle, and finding 
the resultant of each of these two sets of parallel forces, we 
shall obtain, finally, a single force parallel to Y and equal to 

a^xAM, 

and another parallel to OX, equal to 

a^yAM. 



56 APPLIED MECHANICS. 

But, since OZ passes through the centre of gravity of the body, 
we shall have 

%x&M = o and %ybM = o. 

Hence the resultant is in each case, not a single force, but a 
statical couple. Hence, to impart to a body a rotation about 
an axis passing through its centre of gravity requires the action 
of a statical couple ; and conversely, a statical couple so applied 
will cause such a rotation as that described. 

Further discussion of the motion of rigid bodies resulting 
from the action of statical couples is unnecessary for our pres- 
ent purpose, hence we shall pass to the deduction of the fol- 
lowing propositions, viz.: 

PROP. I. Two statical couples in the same plane balance 

each other when they have equal moments, and tend to pro- 

j duce rotation in opposite directions. Let 

J/ F x at a and - i 7 , at b represent one 

a ~ F ^\ couple (left-handed in the figure), and 

F J, /'' \ e let F % at d and — F 9 at e represent the 

yjf ^Tv/jr other (right-handed in the figure), and let 

kxv^ FMb) = nde) ' 

l \ s' ° then will these two couples balance each 

\y' other. 

Proof. — The resultant of F i at a and F t 
FlG< 33, at d will be equal in amount, and directly 

opposed to the resultant of — F 1 at b and — F 9 at e and both 
will act along the diagonal fh of the parallelogram fchg. 
For we have (fg){ab) = (fc){de) t each being equal to the area 
of the parallelogram. 

. F *-Ei . !>-■&. 

hence follows the proposition. 

Hence follows that for a couple we may substitute another 
in the same plane, having the same moment, and tending to- 
rotate the body in the same direction.- 



COUPLES IN THE SAME OR PARALLEL PLANES. 



57 



PROP. II. Two couples in parallel planes balance each 
other when their moments are equal, and the directions in 
which they tend to rotate the body are opposite. 

Let (Fig. 33 (a)) the planes of both couples, be perpendicular 
to OZ. Reduce them both so as to have their arms equal and 
transfer them, each in its own plane, 
till their arms are in the X plane. 
Let ab be the arm of one couple, 
and dc that of the other. Then will 
the two couples form an equilibrate 
system. For the resultant of the 
force at a and that at c acts at e, 
and is twice either one of its com- 
ponents, and hence is equal and di- 




FiG. 33 (a). 



rectly opposed to the resultant of the 
force at b and that at d. 

Hence we may generalize all our propositions in regard to 
the effect of statical couples and we may conclude that — 

In order that two couples may have the same effect, it is 
necessary — 

i°. That they be in the same or pai'allel planes. 

2°. That they have the same moment. 

3°. That they teiid to ca?ise rotation in the same directio7i 
(i.e., both right-handed or both left-handed when looked at from 
the same side). 

It also follows, that, for a given statical couple, we may sub- 
stitute another having the magnitudes of its forces different, 
provided only the moment of the couple remains the same. 

§ 57. Composition of Couples in the Same or Parallel 
Planes. — If the forces of the couples are not 
the same, reduce them to equivalent couples 
having the same force, transfer them to the 
same plane, and turn them so that their arms 
shall lie in the same straight line, as in Fig. 
34 ; the first couple consisting of the force F 
at A and — F at B, and the second of F at B and — F at C. 



c 

\ 



Fig. 34. 



58 



APPLIED MECHANICS. 



The two equal and opposite forces counterbalance each other, 
and we have left a couple with force F and arm 



AC 
Resultant moment 



AB + BC 

F.AC= F(AB) + F(BC), 



Hence : The moment of the couple which is the resultant of 
two or more couples in the same or parallel planes is equal t* 
the algebraic sum of the moments of the component couples. 



EXAMPLES. 

i. Convert a couple whose force is 5 and arm 6 to an equivalent 
couple whose arm is 3. Find the resultant of this and another coupl? 
in the same plane and sense whose force is 7 and arm 8 ; also find the 
force of the resultant couple when the arm is taken as 5. 



2. 



Solution, 



Moment of first couple = 5 x 6 = 30 

When arm is 3, force = $£- =10 

Moment of second couple = 7 X 8 = 56 
Moment of resultant couple = 30 + 56 = 86 
When arm is 5, force = - 8 ¥ 6 - = 17J 

Given the following couples in one plane : — 



Force. 


Arm. 


12 


17 1 


3 


8 


5 


7 


6 


9 


12 


12 


10 


9 


14 


6 J 



Convert to equivalent 
couples having the 
following : — 



Force. 

5 

8 
6 



Arm. 



20 



The first and the last three are right-handed ; the second, third, and 
fourth are left-handed. Find the moment of the resultant couple, and 
also its force when it has an arm n. 



COUPLES IN PLANES INCLINED TO EACH OTHER. 59 

§58. Representation of a Couple by a Line. — From the 
preceding we see that the effect of a couple remains the same 
as long as — 

i°. Its moment does not change. 

2°. The direction of its axis (i.e., of the line drawn perpen- 
dicular to tJie plane of the couple) does not change. 

3 . The direction m which it tends to make the body turn 
{right-handed or left-handed) remains the same. 

Hence a couple may be represented by drawing a line in 
the direction of its axis (perpendicular to its plane), and laying 
off on this line a distance containing as many units of length 
as there are units of moment in the couple, and indicating by a 
dot, an arrow-head, or some other means, in what direction one 
must look along the line in order that the rotation may appear 
right-handed. 

This line is called the Moment Axis of the couple. 

§ 59. Composition of Couples situated in Planes inclined 
to Each Other. — Suppose we have two couples situated 
neither in the same plane nor in parallel planes, and that we 
wish to find their resultant couple. We may proceed as fol- 
lows : Substitute for them equivalent couples with equal arms, 
then transfer them in their own plane respectively to such posi- 
tions that their arms shall _ Rl 
coincide, and lie in the /f\ 
line of intersection of the /If 
two planes. / //f j& 

This having been done, ' V_ %^i ^ s s v 

let OO t (Fig. 35) be the //T^X^^ 

common arm, F and — F / / ' ^\.^jj>* 

the forces of one couple, \ // 01 

F l and —F 1 those of the £ 

other. The forces F and fig. 35 . 

F T have for their resultant R, and — F and —F, have —R,. 

Moreover, we may readily show that R and — R x are equal and 



60 APPLIED MECHANICS. 

parallel, both being perpendicular to 00,. The resultant of 
the two couples is, therefore, a couple whose arm is 00 \ and 
force R } the diagonal of the parallelogram on F and F„ so that 



R = slF 2 + F, 2 + 2i^, cos 0, 

where is the angle between the planes of the couples. Now, 
if we draw from O the line Oa perpendicular to 00, and to F, 
and hence perpendicular to the plane of the first couple, and if 
we draw in the same manner Ob perpendicular to the plane of 
the second couple, so that there shall be in Oa as many units 
of length as there are units of moment in the first couple, and 
in Ob as many units of length as there are units of moment in 
the second couple, we shall have — 

i°. The lines Oa and Ob are the moment axes of the two 
given couples respectively. 

2°. The lines Oa and Ob lie in the same plane with F and 
F J} this plane being perpendicular to 00,. 

3°. We have the proportion 

Oa: Ob = F. 00, :F I .OO i = F\ F„ 

4°. If on Oa and Ob as sides we construct a parallelogram, 
it will be similar to the parallelogram on F and F,. We shall 
have the proportion 

Oc: R ^ Oa: F = Ob: F,; 

and since the sides of the two parallelograms are respectively 
perpendicular to each other, the diagonals are perpendicular to 
each other ; and since we have also 

Oc = R ' 0a and Oa = F. 00, .: Oc = R . OO l} 

it follows that Oc is perpendicular to the plane of the resultant 
couple, and contains as many units of length as there are units 
of moment in the moment of the resultant couple; in other 



COUPLE AND SINGLE FORCE IN THE SAME PLANE. 6l 

words, Oc will represent the moment axis of the resultant 
couple, and we shall have 



or, if we let 



Oc = \l0a 2 + Ob 2 -f 20a . Ob cos aOb; 

Oa = Z, Ob = J/", O = G, aOb = 0, 

G = V'Z 2 + J/ 2 4- 2 LM cos 0. 



This determines the moment of the resultant couple ; and, for 
the direction of its moment axis, we have 



and 



sin aOc = — sin 8 



smbOc — —sin 6. 
G 



Hence we can compound and resolve couples just as we do 
forces, provided we represent the couples by their moment axes 



EXAMPLES. 



i. Given L = 43, M= 15, 6 = 65 ; find resultant couple. 

2. Given L = 40, JZ = 30, = 30 ; find resultant couple. 

3. Given L — 1, M = 5, = 45 ; find resultant couple. 

§ 60. Resultant of a Couple and a Single Force in the 
Same Plane Let M (Fig. 36 or 37) be the moment of the 




l 



Fig. 36. 




J 



given couple, and let OF =z F be the single force. For the 
given couple substitute an equivalent couple, one of whose 
forces is -F at O, equal and directly opposed to the single 



62 



APPLIED MECHANICS. 



force F, these two counterbalancing each other, and leaving 
only the other force of the couple, which is equal and parallel 
to the original single force F, and acts along a line whose 

distance from O is OA — — . Hence — 

F 

The resultant of a single force and a couple in the same plane 
is a force equal and parallel to the original force, having its 
line of direction at a perpendicular distance from the original 
force equal to the moment of the couple divided by the force. 

Fig. 36 shows the case when the couple is right-handed, and 
Fig. 37 when it is left-handed. 

§61. Composition of Parallel Forces in General. — In 
each case of composition of parallel forces (§§ 34, 35, and 36) 
it was stated that the method pursued was applicable to all 
cases except those where 

ZF = o. 



We were obliged, at that time, to reserve this case, because we 
had not studied the action of a statical couple ; but now we will 
ad pt a method for the composition of parallel forces which will 
apply in all cases. 

(a) When all the forces are in one plane. Assume, as we did 
in §35, the axis OY to be parallel to 
the forces ; assume the forces and the 
co-ordinates of their lines of direction, 
as shown in the figure (Fig. 38). Now 
place at the origin O, along OY, two 
equal and opposite forces, each equal to 
F 1 ; then these three forces, viz., F, at 
D, OA, and OB, produce the same effect 
as F l at D alone ; but F 1 at D and OB 
form a couple (left-handed in the figure) 
FlG - 38 - whose moment is —F x x x . Hence the 

force F x is equivalent to — 



Y 












> 






. * 


i 


/ 
t 


k 




. 


F 3 


F,' 


F4 






C 








) 




> 











COMPOSITION OF PARALLEL FORCES. 



63 



i°. An equal and parallel force at the origin, and 

2°. A statical couple whose moment is —F x x x . 

Likewise the force F 2 is equivalent to (i°) an equal and par- 
allel force at the origin, and (2 ) a couple whose moment is 
— F 2 x 2} etc. 

Hence we shall have, if we proceed in the same way with 
all the forces, for resultant of the entire system a single force 

R = ^F along OY } 

and a single resultant couple 

M= -%Fx. 

(Observe that downward forces and left-handed couples are 
to be accounted negative.) 

Now, there may arise two cases. 
i°. When %F= o, and 
2°. When 2F><0. 

Case I. When %F = o, the resultant force along O Y van- 
ishes, and the resultant of the entire system is 
a statical couple whose .moment is 



Case II. When %F > < o, we can reduce 
the resultant to a single force. 

Let (Fig. 39) OB represent the resultant 
force along O Y, R = %F With this, compound 
the couple whose moment is M — —%Fx, and 
we obtain as resultant (§ 60) a single force 



R = %F, 



^x 



Fig. 39. 



whose line of action is at a perpendicular distance from OY 
equal to 

%Fx 



AO = x r = 



XF 



64 



APPLIED MECHANICS. 




Fig. 40. 



(fr) When the forces are not confined to one plane. Assume, 
as before (Fig. 40), OZ parallel to the forces, and let F acting 

through A be one of the given 
forces, the co-ordinates of A be- 
ing x and y. Place at two equal 
and opposite forces, each equal to 
F, and also at B two equal and 
opposite forces, each equal to F. 
These five forces produce the 
same effect as F alone at A, and 
they may be considered to con- 
sist of — 
i°. A single force F at the origin. 

2 . A couple whose forces are F at B and — F at (9, and 
whose moment is — Fx acting in the y plane. 

3 . A couple whose forces are F at A and — F at B, and 
whose moment is Fy acting in the x plane. Treating each of 
tne forces in the same way, we shall have, in place of the entire 
system of parallel forces, the following forces and couples : — 
i°. A single force R = %F along OZ. 
2°. A couple My = —%Fx in the y plane. 
3 . A couple M x — +%Fy in the x plane. 
Now, there may be 
two cases : — 

i°. When 5^X0. 
2 . When %F = o. 

Case I. When $F> 
< o, we can reduce to a 
single resultant force 
having a fixed line of 
direction. Lay off (Fig. 
41) along OZ,OH=%F. 
Combining this with the first of the above-stated couples, we 




Fig. 41. 



COMPOSITION OF PARALLEL FORCES. 



65 



%Fx 
obtain a force R = %F at A, where OA = — — = x r . Then 

^F 

combine with this resultant force R = %F at A } the second 

couple, and we shall have as single resultant of the entire 

system a single force 

R = %F 
acting through B> where 



AB 



)'r --= 



XFy 



Hence the resultant is a force whose magnitude is 

R = IF, 



the co-ordinates of its line of direction being 



%Fx 



2F' Jr %F' 



Case II. When %F*= o, there is no single resultant force ; 
but the system reduces to two couples, one in the x plane and 
one in the y plane, and these two can be reduced to one single 
resultant couple. (Observe that couples are to be accounted 
positive when, on being looked at by the observer from the posi- 
tive part of the axis towards 
the origin, they are right- 
handed ; otherwise they are 
negative.) 

The moment axis of the 
couple in the x plane will 
be laid off on the axis OX 
from the origin towards the 
positive side if the moment 
is positive, or towards the 
negative side if it is nega- fig. 42- 

tive, and likewise for the couple in the y plane. 




— — *x 



66 



APPLIED MECHANICS. 



Hence lay off (Fig. 42) OB = M x , OA = M y , and by 
completing the rectangle we shall have OD as the moment 
axis of the resultant couple ; hence the resultant couple lies 
in a plane perpendicular to OD, and its moment bears to 
OD the same ratio as M x bears to OB. 

Hence we may write 

OD = M r = \fM x 2 + M/, 
cos DOX = M* = cos0. 

M r 

If My had been negative, we should have OB as the moment 
axis of the resultant couple. 

EXAMPLES. 





P. 


X. 


y- 




/?. 


#. 




I. 


5 


4 


3 


2. 


5 


-4 


3 




3 


2 


1 




— 2 


2 


— 1 


WL . 


1 


3 


5 




-3 


3 


5 



Find the resultant in each example. 

§62. Resultant of any System of Forces acting at Dif- 
ferent Points of a Rigid 
Body, all situated in One 
Plane. — Let CF = F (Fig. 
43) be one of the given forces. 
Let all the forces be referred 
to a system of rectangular 
axes, as in the figure, and let 
a = angle made by F with 
fig. 43. OX, etc. Let the co-ordi- 

nates of the point of application of F be AO — x, BO =zy. 



SYSTEM OF FORCES ACTING ON RIGID BODY. 6 J 

We first decompose CF = F into two components, parallel 
respectively to OX and Y These components are 

CD — T^cosa, CE — i^sina. 

Apply at O in the line O Y two equal and opposite forces, each 
equal to Fsm a, and at in the line OX two equal and opposite 
forces, each equal to Fcosa. Since these four are mutually 
balanced, they do not alter the effect of the single force; and 
hence we have, in place of .Fat C, the six forces CD, OM, OK, 
CE, ON, OG. Of these six, CE and OG form a couple whose 
moment is 

— {Fsma)x = — Fx€wia, 

CD and OK form a couple whose moment is 
{Fco%a)y — Fy cos a. 

These two couples, being in the same plane, give as result- 
ant moment their algebraic sum, or 

F{y cos a — x sin a) . 

We have, therefore, instead of the single force at C, the follow- 
ing:— 

i°. OM = Fcos a along OX. 

2°. ON— Fsinaalong OY. 

3°. The couple M = F(y cos a — x sin a) in the given plane. 

Decompose in the same way each of the given forces ; and 
we have, on uniting the components along OX, those along O Y, 
and the statical couples respectively, the following : — 

i°. A resultant force along OX, R x = %Fcos a. 

2°. A resultant force along OY, R y = %Fsma. 

3°. A resultant couple in the plane, whose moment is 

M = XFLy cos a — x sin a}. 



68 



APPLIED MECHANICS. 



This entire system, on compounding the two forces at O x 
reduces to 



i°. R = \lR x * + Rf = y/(^Fcosa)» + (XFsina)*; 
making with OX an angle a r , where 

%F cos a 



COS (V = 



R 



2°. A resultant couple in the same plane, whose moment is 

M — %F(y cos a — x sin a) . 

Now compound this resultant force and couple, and we have, 

for final resultant, a single 
force equal and parallel to 
R, and acting along a line 
whose perpendicular dis- 
tance from O is equal to 

M 
R' 





Y 

F 




F 










G 


— ' 

t R 


C 


D 
X 


" \ 






L O 


E 


\- 


\ 



Fig. 44. 

Suppose (Fig. 44) the force 

OB = 2^ cos a, 
OA = %F sin a, 



The equation of this line 
may be found as follows : 



OR = \/($R cos a) 2 -f (5^ sin a) 2 ; 

and let us suppose the resultant couple to be right-handed, and 
let 

OM=M. 
R 

then will the line ME parallel to OR be the line of direction 
of the single resultant force. 



CONDITIONS OF EQUILIBRIUM. 69 

Assuming the force R to act at any point C (x r , y r ) of this' 
line, if we decompose it in the same way as we did the single 
forces previously, we obtain — 

i°. The force R cos a r — XFcos a along OX. 

2°. The force R sin a r = X^sina along OY. 

3 . The couple R(y r cos a r — x r sin o^). 

Hence we must have 

R(y r cos a r — ^ r sina r ) = "%R(ycosa — 3: sin a) = M. 
Hence for the equation of the line of direction we have 

y r cos a r — x r sm cv = — . (1) 

^ R 

Another form for the same equation is 

jv(XFcosa) - x r (2,Fsma) = M. (2) 

§ 63. Conditions of Equilibrium. — If such a set of forces 
be in equilibrium, there must evidently be no tendency to trans- 
lation and none to rotation. Hence we must have 

R = o and M = o. 

Hence the conditions of equilibrium for any system of forces 
in a plane are three ; viz., — 

2i^cosa = o, 1,F sin a = o, 2LF(jycosa — ^sina) = o. 

Another and a very convenient way to state the conditions of 
equilibrium for this case is as follows : — 

If the forces be resolved into components along two directions 
at right angles to each other, then the algebraic sum of the com- 
ponents along each of these directions must be zer&, and th*. 
algebraic sum of the moments of the forces about any axis pe/* 
pendicular to the plane of the forces must equal zero. 



;o 



APPLIED MECHANICS. 



i. Given 



2. Given 







EXAMPLES. 


p. 


X. 


y. a. 


5 


3 


2 3 1 ' 


IO 


i 


3 40 


-7 


4 


2 54 c 


P. 


#. 


?. a. 


12 


27 


3 15 


4 


i3 


-5 30' 


8 


-5 


-4 45 



Find the resultant, and 
the equation of its 
line of direction. 

I Find the resultant, and 
the equation of its 
line of direction. 



I 



§64. Resultant of any System of Forces not confined 

to One Plane. — Suppose we 
have a number of forces applied 
at different points of a rigid 
body, and acting in different 
directions, of which we wish to 
find the resultant. Refer them 
all to a system of three rect- 
angular axes, OX, OY, OZ 
(Fig. 45). Let PR — F be 
one of the given forces. Re- 
solve it into three components, 
PK, PH, and PG, parallel 
Let 




Fig. 45. 

respectively to the three axes. 



RPK 



RPH = /?, RPG = y. 



Let OA = x, OB = y, OC = z, be the co-ordinates of the 
point of application of the force F. Now introduce at B and 
also at O two forces, opposite in direction, and each equal to PK. 
We now have, instead of the force PK, the five forces PK, BM, 
BN, OS, and OT. The two forces PK and BN form a couple 
in the y plane, whose axis is a line parallel to the axis OY, and 
whose moment is {PK)(EB) = (Fco$>a)z = Fz cos a. The 



FORCES NOT CONFINED TO ONE PLANE. fl 

forces BM and OT form a couple in the z plane, whose moment 

is 

(BM)(OB) = -Fy cos a. 

Now do the same for the other forces PH and PG, and we shall 
finally have, instead of the force PR, three forces, 

F cos a, F cos ft, F cos y, 

acting at O in the directions (XY, OY, and <9Z respectively, 
together with six couples, two of which are in the x plane, two 
in thej*/ plane, and two in the z plane. 

They thus form three couples, whose moments are as fol- 
lows : — 

Around OX, F(y cos y — z cos /?) j 

Around OY, F(z cos a — x cos y); 

Around OZ, F(x cos /? — y cos a) . 

Treat each of the given forces in the same way, and we shall 
have, in place of all the forces of the system, three forces, 

%F cos a along OX, 
IF cos j3 along OY, 
^Fcosy along OZ; 

and three couples, whose moments are as follows : — 

Around OX, M x = XF(y cos y — z cos /?) ; 
Around OY, M y — %F(z cos a — a: cosy); 
Around OZ, M z = ^F(xcos/3 — j'cosa). 

The three forces give a resultant at O equal to 



R = V(XFcosa) 2 + (ZFcosP) 2 4- (2^ cosy) 2 , (i) 

™. 2^ cos a ^ rtc , ^ ^F cos 13 . „ XFcosy / x 

008 0^. = — — , cos ft. = — -£--£, cosy r = — ^-A ( 2 ) 



72 



APPLIED MECHANICS. 



For the three couples we have as resultant 



M= s/M x 2 -f M/ + M z 2 , 

x J/. J/y M z 

cosA=— , co S/ x = -^, cos " = ^ 



(3) 

(4) 



A, /x, and v being the angles made by the moment axis of the 
resultant couple with OX, O Y, and OZ respectively. 

Thus far we have reduced the whole system to a single result- 
ant force at the origin, and a couple. Sometimes we can reduce 

the system still farther 
and sometimes not. The 
following investigation will 
show when we can do so. 
Let (Fig. 46) OP — R be 
the resultant force, and 
OC =M the moment axis 
of the resultant couple. 
Denote the angle between 
them by 6 (a quantity thus 
far undetermined). Pro- 
ject OP = R on OC. Its 
projection will be OD = RcosO; then project, in its stead, the 
broken line OABP on OC. By the principles of projections, 
the projection of this broken line will equal OD. 

Now OA y AB, and BP are the co-ordinates of P, and make 
with OC the same angle as the axes OX, OY, and OZ ; i.e., 
A, fx, and v respectively : hence the length of the projection is 

OA cos X + AB cos /x -f BPcosv. 

But 

OA = j^coscv, AB = R cos /3 r , BP = ^cosy^. 

Hence 

R cos 6 — R cos or cos A 4- R cos f} r cos /x + R cos y r cos v 
COS0 = COS Or cos A -f- COS /3 r COS /x -|- cosy r cosv. (5) 




Fig. 46. 



CONDITIONS OF EQUILIBRIUM. 73 

This enables us to find the angle between the resultant force 
and the moment axis of the resultant couple. 

The following cases may arise : — 

i°. When cos = o, or — 90 , the force lies in the plane 
of the couple, and we can reduce to a single force acting at a 

distance from O equal to — , and parallel to R at O. 

R 

2°. When cos == I, or = o, the moment axis of the 
couple coincides in direction with the force : hence the plane 
of the couple is perpendicular to the force, and no farther 
reduction is possible. 

3 . When 6 is neither o° nor 90 , we can resolve the couple 
M into two component couples, one of which, McosO, acts in a 
plane perpendicular to the direction of R, and the other, M sin 0, 
acts in a plane containing R. The latter, on being combined 
with the force R at the origin, gives an equal and parallel force 
whose line of action is at a distance from that of R at 0, equal 
to 

MsinO 



4 . When M = o, the resultant is a single force at O. 

5°. When R = o, the resultant is a couple. 

§65. Conditions of Equilibrium. — To produce equilibrium, 
we must have no tendency to translation and none to rotation. 
Hence we must have 

R = o and M = o. 

Hence we have, in general, six conditions of equilibrium ; viz.,— 

Si 7 cos a = o, 2LFcos/2 = o, XF cosy = o. 
M x = o, My s o, M z — o. 



74 APPLIED MECHANICS. 



EXAMPLES. 

i. Prove that, whenever three forces balance each other, they must 
lie in one plane. 

2. Show how to resolve a given force into two whose sum is given, 
the direction of one being also given. 

3. A straight rod of uniform section and material is suspended by two 
strings attached to its ends, the strings being of given length, and attached 
to the same fixed point : find the position of equilibrium of the rod. 

4. Two spheres are supported by strings attached to a given point, 
and rest against each other : find the tensions of the strings. 

5 . A straight rod of uniform section and material has its ends resting 
against two inclined planes at right angles to each other, the vertical 
plane which passes through the rod being at right angles to the line of 
intersection of the two planes : find the position of equilibrium of the 
rod, and the pressure on each plane, disregarding friction. 

6. A certain body weighs 8 lbs. when -placed in one pan of a false 
balance of equal arms, and 10 lbs. in the other : find the true weight of 
the body. 

7. The points of attachment of the three legs of a three-legged table 
are the vertices of an isosceles right-angled triangle ; a weight of 100 lbs. 
is supported at the middle of a line joining the vertex of one of the acute 
angles with the middle of the Opposite side : find the pressure upon 
each leg. 

8. A heavy body rests upon an inclined plane without friction : find 
the horizontal force necessary to apply, to prevent it from falling. 

9. A rectangular picture is supported by a string passing over a 
smooth peg, the string being attached in the usual way at the sides, but 
one-fourth the distance from the top : find how many and what are the 
positions of equilibrium, assuming the absence of. friction. 

10. Two equal and weightless rods are jointed together, and form a 
right angle ; they move freely about their common point : find the 
ratio of the weights that must be suspended from their extremities, that 
one of them may be inclined to the horizon at sixty degrees. 

11. A weight of 100 lbs. is suspended by two flexible strings, one 
of which is horizontal, and the other is inclined at an angle of thirty 
degrees to the vertical : find the tension in each string. 



D YNAMICS. — DEFINITIONS. ?$ 



CHAPTER II. 

DYNAMICS. 

§ 66. Definitions. — Dynamics is that part of mechanics 
which discusses the forces acting, when motion is the result. 

Velocity, in the case of uniform motion, is the space passed 
over by the moving body in a unit of time ; so that, if s repre- 
sent the space passed over in time t, and v represent the velocity, 
then 

Velocity, in variable motion, is the limit of the ratio of the 
space (As) passed over in a short time (At), to the time, as the 
latter approaches zero : hence 

ds 

v = --. 
dt 

Acceleration is the limit of the ratio of the velocity \A^) im- 
parted to the moving body in a short time (At), to the time, as 
the time approaches zero. Hence, if a represent the accelera- 
tion, 



_ dv = d \dt) __ d^s 
dt dt dt* 



y6 APPLIED MECHANICS. 

§ 67. Uniform Motion. — In this case the acceleration is 
zero, and the velocity is constant ; and we have the equation 

s = vt. 

§ 68. Uniformly Varying Motion. — In this case the ac- 
celeration is constant : hence a is a constant in the equation 

d 2 s 

— - = a, 

dt 2 

and we obtain by one integration 

ds . . 

v = — = at -f- c, 
dt 

where c is an arbitrary constant : to determine it we observe, 
that, if v Q represent the value of v when / = o, we shall have 

v = o ■+■ c 

.*. c = v 

ds , , 

•*• v = — = at + v , 
at 

and by another integration 

s = %at 2 4- v Q t 3 

jviiers s is the space passed over in time / ; the arbitrary con- 
stant vanishing, because, when t = o, s is also zero. 

§ 69. Measure of Force. — It has already been seen, that, 
when a body is either at rest or moving uniformly in a straight 
line, there are either no forces acting upon it, or else the forces 
acti" j upon it are balanced. If, on the other hand, the motion 
of i>e body is rectilinear, but not uniform, the only unbalanced 
force acting is in the direction of the motion, and equal in mag- 
nitude to the momentum imparted in a unit of time in the direc- 
tion of the motion, or, in other words, to the limit of the ratio 
of the momentum imparted in a short time (A*), to the time, as 
the latter approaches zero. 



MECHANICAL WORK. — UNIT OF WORK. J J 

Thus, if F denote the force acting in the direction of the 
motion, m the mass, and a the acceleration, we shall have 

c, dv d 2 s (i) 

F — ma = m — = m — . v ' 

dt dt 2 

From (i) we derive 

mdv — Fdt; (2) 

and, if v Q be the velocity of the moving body at the time when 
/ = t Q , and v x its velocity when / = t a we shall have 



Fdt 
or 



mdv =11 

m(v, —v ) = i Fdt; (3) 



to 

or, in words, the momentum imparted to the body during the 
time t = (/ x — t Q ) by the force F y will be found by integrating 
the quantity Fdt between the limits t x and t Q . 

§ 70. Mechanical Work. — Whenever a force is applied to 
a moving body, the force is either used in overcoming resist- 
ances (i.e., opposing forces, such as gravity or friction), and 
leaving the body free to continue its original motion undis- 
turbed, or else it has its effect in altering the velocity of the 
body. In either case, the work done by the force is the prod- 
uct of the force, by the space passed through by the body f.n 
the direction of the force. 

Unit of Work. — The unit of work is that work which is 
done when a unit of force acts through a unit of distance in 
the same direction as the force ; thus, if one pound and one 
foot are our units of force and length respectively, the unit of 
work will be one foot-pound. 

If a constant force act upon a moving body in the direction 
of its motion while the body moves through the space s f the 
work done by the force is 

Fs; 



78 APPLIED MECHANICS. 

and this, if the force is unresisted, is the energy, or capacity for 
performing work, which is imparted to the body upon which the 
force acts while it moves through the space s. 

Thus, if a io-pound weight fall freely through a height of 
5 feet, the energy imparted to it by the force of gravity during 
this fall is 10 X 5 = 50 foot-pounds, and it would be necessary 
to do upon it 50 foot-pounds of work in order to destroy the 
velocity acquired by it during its fall. If, on the other hand, 
the force is a variable, the amount of work done in passing 
over any finite space in its own direction will be found by in- 
tegrating, between the proper limits, the expression 

fFds. 

The power which a machine exerts is the work which it 
performs in a unit of time. 

The unit of power commonly employed is the horse-power y 
which in English units is equal to 33000 foot-pounds per 
minute, or 550 foot-pounds per second. 

§71. Energy. — The energy of a body is its capacity for 
performing work. 

Kinetic or Actual Energy is the energy which a body pos- 
sesses in virtue of its velocity ; in other words, it is the work 
necessary to be done upon the body in order to destroy its 
velocity. This is equal to the work which would have to be 
done to bring the body from a state of rest to the velocity with 
which it is moving. Assume a body whose mass is m y and sup- 
pose that its velocity has been changed from v Q to v v Then if 
F be the force acting in the direction of the motion, we shall 
have, from equation (2), § 69, that 

Fvdt = mvdv; (1) 

but 

vdt — ds 

/. Fds = mvdv. (2) 



AT WOOD'S MACHINE. 79 

Hence, by integration, 

Jmvdv = I Fds 
Vq U 

.-. \m(v* - v Q 2 ) = fFds; (3) 

but fFds is the work that has been done on the body by the 
force, and the result of doing this work has been to increase 
its velocity from v to v v It follows, that, in order to change 
the velocity from v Q to v„ the amount of work necessary to per- 
form upon the body is 

\m(y* - v >) = i —(vS - vi). (4) 

g 

If v Q = o, this expression becomes 

*■*', or ** (S) 

which is the expression for the kinetic energy of a body of mass 
m moving with a velocity v v 

§ 72. Atwood's Machine. — A particular case of uniformly 
accelerated motion is to be found in Atwood's machine, in which 
a cord is passed over a pulley, and is loaded with unequal weights 
on the two sides. Were the weights equal, there would be no 
unbalanced force acting, and no motion would ensue ; but when 
they are unequal, we obtain as a result a uniformly accelerated 
motion (if we disregard the action of the pulley), because we 
have a constant force equal to the difference of the two weights 
acting on a mass whose weight is the sum of the two weights. 
Thus, if we have a 10-pound weight on one side and a 5-pound 
weight on the other, the unbalanced force acting is 

F = 10 - 5 = 5 lbs. 



80 APPLIED MECHANICS. 



The mass moved is M = ^t-JL . hence the resulting ac- 

g 
celeration is 



m 



§ 73. Normal and Tangential Components of the Forces 
acting on a Heavy Particle. — If a body be in motion, either 
in a straight or in a curved line, and if at a certain instant all 
forces cease acting on it, the body will continue to move at a 
uniform rate in a straight line tangent to its path at that point 
where the body was situated when the forces ceased acting. 

If an unresisted force be applied in the direction of the 
body's motion, the motion will still take place in the same 
straight line ; but the velocity will vary as long as the force 
acts, and, from what we have seen, the equation 

F = m — ( 1 ) 

dt 2 V ; 

will hold. 

If an unresisted force act in a direction inclined to the 
body's motion, it will cause the body to change its speed, and 
also its course, and hence to move in a curved line. Indeed, 
if a force acting on a body which is in motion be resolved into 
two components, one of which is tangent to its path and the 
other normal, the tangential component will cause the body to 
change its speed, and the normal component will cause it to 
change the direction of its motion. 

The measure of the tangential component is, as we have 
seen, 

77 d 2 s 

dt* 

and we will proceed to find an expression for the normal com- 
ponent otherwise known as the Deviating Force. For this 



CENTRIFUGAL FORCE. 



Si 



purpose we may substitute, for a small portion of the curve, a 
portion of the circle of curvature ; hence we will proceed to 
find an expression for the centrifugal force of a body which 
moves uniformly with a velocity z/ina circle whose radius is r. 



CENTRIFUGAL FORCE. 



Let AC (Fig. 47) be the space described in the time At. 



Then we have 



AC = vAt. 



The motion AC may be approximately consid- 
ered as the result of a uniform motion 

AB == vAt nearly, 

and a uniformly accelerated motion 

BC = \a(At) 2 = s, 

where a = acceleration due to centrifugal force. But 

(AB) 2 = BC .BD, 

{z>At) 2 = ia(At) 2 (2r + s), 

AO = OC = r 
.*. v 2 = \a{2r -f j") approximately 




Pig. 47. 



or 
where 



# = - 



2Zr 



approximately. 



2r + j- 

For its true value, pass to the limit where s = o. 

Hence we have, for the acceleration due to the centrifugal 
force, the expression 

r 



Hence the centrifugal force is equal to 



p _ mv 2 



(«) 



82 APPLIED MECHANICS. 



DEVIATING FORCE. 

If a body is moving in a curved path, whether circular or 
not, and the unbalanced force acting on it be resolved into tan- 
gential and normal components, the tangential component will 
be, as has already been seen, 

d 2 s 

dt 2 

and the normal component will be 

mv 2 _ m/dsV 
r ~~ ~r\dt) ' 

where r is the radius of curvature of the path at the point in 

question. 



RESULTANT FORCE. 



Hence it follows that the entire unbalanced force acting on 
the body will be 



. // d 2 s\ 2 , fm ds 2 \ 2 



or 



*-nW+ ?(£)' » 



§ 74. Components along Three Rectangular Axes of the 

Velocities of, and of the Forces acting on, a Moving 

ds 
Body. — If we resolve the velocity— into three components 

dt 

along OX, OY, and OZ, we shall have, for these components. 

respectively, 

d A & and ^; 
dt dt dt 

this being evident from the fact that dx, dy, and dz are respec- 



COMPONENTS OF VELOCITIES AND FORCES. 8$ 

tively the projections of ds on the axes OX, OY, and OZ ; and, 
from the differential calculus, we have 



!-#)'+©' + (S)' 



On the other hand, 



d A &, and * 
dt dt dt 



ds . 
are not only the components of the velocity — in the directions 

dt 

OX, OY, and OZ, but they are also the velocities of the body 

in these directions respectively. 

Now, the case of the accelerations is different ; for, while 



d 2 x d 2 y , d 2 z 

——, -f-, and — - 
dt 2 dt 2 dt 2 



are the accelerations in the directions OX, OY, and OZ respec- 
tively, they are not the components of the acceleration 

d^s 
dt 2 

along the three axes. 

That they are the former is evident from the fact that — , 

dt 

-*-, and — are the velocities in the directions of the axes, and 
dt dt 

——, —?-, — — are their differential co-efficients, and hence repre- 
dt 2 dt 2 dt 2 * 

sent the accelerations along the three axes. But if we consider 

the components of the force acting on the body, we shall have 



84 APPLIED MECHANICS. 

for its components along OX, O Y, and OZ, if a, /?, and y are 
the angles made by F with the axes respectively, 

c, </ 2 .# £, d 2 y r, d 2 z 

dt 2 r dt 2 r dt 2 



and we found (§ 73) for F, the value 



—J®l + i@f- «■> 



Hence, equating these values of F, and simplifying, we shall 
have the equation 

(d 2 x\* (dy\ 2 /d^z\ 2 = fd^sY , i^MV 
\dt 2 ) \dt 2 ) \dt 2 ) ' \dt 2 ) r\dt) 

Hence it is plain that — , — ^-, and -— - can only be the com- 
F dt 2 dt 2 dt 2 J 

ponents of the actual acceleration 

d 2 s 
dt 2 

when the last term — ( — ) vanishes, or when r = 00 , i.e., when 
r 2 \dt) 

the motion is rectilinear. 

Moreover, we have the two expressions (1) and (2) for the 
force acting upon a moving body. 

The truth of the proposition just proved may also be seen 
from the following considerations :- — 

If a parallelopiped be constructed with the edges 

dx dy dz 
It' dt' dt' 



CENTRIFUGAL FORCE OF A SOLID BODY. 8$ 

; ■ 

the diagonal will be the actual velocity 

ds 
dt' 

and will, of course, coincide in direction with its path. 

On the other hand, if a parallelopiped be constructed with 

the edges 

d 2 x d 2 y d^z 
dt 2 ' dt 2 ' dt 2 ' 

its diagonal must coincide in direction with the force 



'"tf©*i© 



and can coincide in direction with the path, and hence with the 
actual acceleration 

d 2 s 

dt 2 ' 

only when the force is tangential to the path, and hence when 
the motion is rectilinear. 

§75. Centrifugal Force of a Solid Body. — When a solid 
body revolves in a circle, the resultant centrifugal force of the 
entire body acts in the direction of the perpendicular let fail 
from the centre of gravity of the body on the axis of rotation, 
and its magnitude is the same as if its entire weight were con- 
centrated at its centre of gravity. 

Proof. — Let (Fig. 48) the angular velocity = a, and the *cta* 
weight = W. Assume the axis of rotation perpendicular t 
the plane of the paper and passing through 
O ; assume, as axis of x, the perpendicular 
dropped from the centre of gravity upon 
the axis of rotation. The co-ordinates of 
the centre of gravity will then be (x of jy Q ), 
and y Q will be equal to zero. 

If, now, P be any particle of weight w, 
where r =■ perpendicular distance from P on axis of rotation, 




B6 APPLIED MECHANICS. 

and x = OA, y = AP, we shall have for the centrifugal force 
of the particle at P 

W 2 

£ 

but if we resolve this into two components, parallel respectively 
to OX and OY, we shall have for these components 

/W . \X a 2 , /W , \ V a 2 

(~a 2 r\- = —wx and [-a 2 ry- = -wy, 
Xg / r g \g / r g 

and, for the resultant for the entire body we shall have, parallel 
to OX, 

F x = -2wx = -Wx oy (i) 

g g 

and 

F y = —2wy — — Wy = o. (2) 

g g 

Hence the centrifugal force of the entire body is 

F=-fVx Q ; (3) 

g 

ani if we let v Q = ax = linear velocity of the centre of gravity, 

we have 

Wv 2 



P = 



g*o 



wm.h is the same as though the entire weight of the body 
veic concentrated at its centre of gravity. 



EXAMPLES. 

0. A 10-pound weight is fastened by a rope 5 feet long to the 
centre, aroun 1 which it revolves at the rate of 200 turns per minute ; 
rird the pull on the cord. 

2. A locomotive weighing 50000 lbs., whose driving-wheels weigh 
t©c lbs., is running at 60 miles per hour, the diameter of the drivers 



UNIFORMLY VARYING RECTILINEAR MOTION. 87 



being 6 feet, and the distance from the centre of the wheel to the centre 
of gravity of the same being 2 inches (the drivers not being properly 
balanced) ; find the pressure of the locomotive on the track (a) when 
the centre of gravity is directly below the centre of the wheel, and (b) 
when it is directly above. 

3. Assume the same conditions, except that the distance between 
centre of the wheel and its centre of gravity is 5 inches instead of 2. 

§76. Uniformly Varying Rectilinear Motion. — We have 
already found for this case (§ 68) the equations 

d 2 s 

— - = a = a constant, 

dt 2 ' 

ds 

-- = v = v + at, 

at 

s = vj + \at 2 ; 

and we may write for the force acting, which is, of course, coin- 
cident in direction with the motion, 

zr d 2 s 

* = m—- = ma = a constant. 
at 2 

§ 77. Motion of a Body acted on by the Force of Gravity 
only. — A useful special case of uniformly varying motion is 
that of a body moving under the action of gravity only. 

The downward acceleration due to gravity is represented by 
g feet per second, the value of g varying at different points on 
the surface of the earth according to the following law : — 

g = ^1(1 — 0.00284 cos 2A.)(i — — \ feet per second, 

where 

g, = 32.1695 feet, 

A = latitude of the place, 

h — its elevation above mean sea-level in feet, 

R = 20900000 feet. 



88 APPLIED MECHANICS. 

If, now, we represent by h the height fallen through by a 
descending body in time t, we shall have the equations, 

v = v + gt, 
h = v Q t + \gi 2 , 

where v is the initial downward velocity. 

If, on the other hand, we represent by v Q the initial upward 
velocity, and by h the height to which the body will rise in 
time / under the action of gravity only, we must write the equa- 
tions 

v = v - gt, 

h = v Q t - \gt\ 

When v Q = o, the first set of equations gives 

v = gt, 
h = igt 2 , 

which express the law of motion of a body starting from rest 
and subject to the action of gravity only. 

Eliminate t between these equations, and we shall have 

v 1 = 2gh /. v = S2gh, 
or 

h is called the height due to the velocity v, and represents the 
height through which a falling body must drop to acquire the 
velocity v ; and 

v = \2gh 



UNRESISTED PROJECTILE. 89 



is the velocity which a falling body will acquire in falling 
through the height h. Thus, if a body fall through a height of 
50 feet, it will, by that fall, acquire a velocity of about 



^2(32.1) (cj ) = ^3216.66 = 56.7 feet per second. 

Again : if a body has a velocity of 40 feet per second, we shall 
have 

v 1 1600 



2g 64.3 



= 24.8 feet ; 



and we say that the body has a velocity due to the height 24.8 
feet, i. e., a velocity which it would acquire by falling through a 
height of 24.8 feet. 



EXAMPLES. 

1. A stone is dropped down a precipice, and is heard to strike the 
bottom in 4 seconds after it started : how high is the precipice ? 

2. How long will a stone, dropped down a precipice 500 feet high, 
take to reach the bottom? 

3. What will be its velocity just before striking the ground? 

4. A body is thrown vertically upwards with a velocity of 100 feet 
per second ; to what height will it rise ? 

5. A body is thrown vertically upwards, and rises to a height of 50 
feet. With what velocity was it thrown, and how long was it in its 
ascent ? 

6. What will be its velocity in its ascent at a point 15 feet above 
the point from which it started, and what at the same point in its 
descent ? 

§ 78. Unresisted Projectile. — In the case of an unresisted 
projectile, we have a body on which is impressed a uniform 



9° 



APPLIED MECHANICS. 



motion in a certain direction (the direction of its initial motion), 
and which is acted on by the force of gravity only. 

Let OPC be 
the path (Fig. 49), 
OA the initial di- 
rection, and v Q the 
initial velocity, and 
the angle AOX — 

e. 

Then we shall 
- have, for the hori- 

TIG. 49. ' 

zontal and vertical 
components of the unbalanced force acting, when the projectile 
is at P (co-ordinates x and y), 




m — = along OX, and m — 



dt 2 



dt 2 



mg = — W along OY. 



Hence 



d*x 
~di 2 



= o, 



(1) 



d 2 y 
~dP 



~g- 



(2) 



Integrating, and observing, that, when t = o, the horizontal 
and the vertical velocities were respectively v Q cos 6 and v Q sin 0, 
we have 

= v cos 0, (3) 



dt 
dy 
It 



v sin — gt 



(4) 



These equations could be derived directly by observing that 
the horizontal component of the initial velocity is v cos 0, and 
that this remains constant, as there is no unbalanced force act- 
ing in this direction, also that v o s'm0 is the initial vertical 
velocity ; and, since the body is acted on by gravity only, this 
velocity will in time / be decreased by gt. 



UNRESISTED PROJECTILE. 9 1 

Integrating equations (3) and (4), and observing that for 
/ — o, x and y are both zero, we obtain 

x — v cos 6 J, (5) 

y = v sin 6 J - %g* 2 - (fi) 

Eliminate /, and we have 

v = *tanfl &*- (7) 

as the equation of the path, which is consequently a parabola. 

Equations (1), (2), (3), (4), (5), (6), and (7) enable us to solve 
any problem with reference to an unresisted projectile. 

Equation (7) may be written 

/ _ Vq 2 sin 2 fl \ _ g _ / _ v 2 sin fl cos fl y , g . 

V 2^ / ~ 2V 2 cos 2 fl \ ^ . / ^ ' 

which gives for the co-ordinates of the vertex 

_ v 2 sin 2 fl __ v 2 sin fl cos 
y t — , x x — • 

2g g 

EXAMPLES. 

i. An unresisted projectile starts with a velocity of 100 feet per 
second at an upward angle of 30 to the horizon ; what will be its velocity 
when it has reached a point situated at a horizontal distance of iooq teet 
from its starting-point, and how long will be required for it te re?c*» 
that point ? 

Solution. 

v Q = 100, fl = 30 , v Q cos fl = 86.6, v sic $ = 50^ 
g = 32.16. 
Equation (5) gives us 

1000 = 86.6 t 

•\ /= = 11.51; seconds. 

86,6 



92 



APPLIED MECHANICS. 



e/ sin# ~ gt = 50 — 371.5 = -321.5, 



= V^(86.6) 2 + (321.5) 2 = V75 00 + 103362 = 333. 



Hence the point in question will be reached in \\\ seconds after start- 
ing, and the velocity will then be 333 feet per second. 

2. An unresisted projectile is thrown upwards from the surface of 
the earth at angle of 39 to the horizontal : find the time when it will 
reach the earth, and the velocity it will have acquired when it reaches 
the earth, the velocity of throwing being 30 feet per second. 

3. A 10-pound weight is dropped from the window of a car when 
travelling over a bridge at a speed of 25 miles an hour. How long will 
it take to reach the ground 100 feet below the window, and what will be 
the kinetic energy when it reaches the ground ? 

4. With what horizontal velocity, and in what direction, must it be 
thrown, in order that it may strike the ground 50 feet forward of the 
point of starting ? 

5. Suppose the same 10-pound weight to be thrown vertically up- 
wards from the car window with a velocity of 100 feet a minute, how 
long will it take to reach the ground, and at what point will it strike the 
ground ? 

§ 79. Motion of a Body on an Inclined Plane without 

Friction. — If a body move on 
an inclined plane along the line 
of steepest descent, subject to 
the action of gravity only, and 
if we resolve the force acting 
on it (i.e., its weight) into two 
components, along and perpen- 
dicular to the plane respec- 
tively, the latter component 
will be entirely balanced by 
the resistance of the plane, 

and the former will be the only unbalanced force acting on 

the body. 




MOTION OF A BODY ON AN INCLINED PLANE. 93 

Suppose a body whose weight is represented (Fig. 50) by 
HF = W to move along the inclined path AB under the action 
of gravity only. Let 6 be the inclination of AB to the horizon. 
Resolve W'mto two components, 

HD=Ws\nO, and BE = W cos 0, 

respectively parallel and perpendicular to the plane. The 
former is the only unbalanced force acting on the body, and 
will cause it to move down the plane with a uniformly accel- 
erated motion ; the acceleration being 

^sin . a , . 

= g sin 0. (1) 



(f) 



If the body is either at rest or moving downwards at the 
beginning, it will move downwards ; whereas, if it is first mov- 
ing upwards, it will gradually lose velocity, and move upwards 
more slowly, until ultimately its upward velocity will be de- 
stroyed, and it will begin moving downwards. 

The equations for uniformly varying motion are entirely 
applicable to these cases. Thus, suppose that the body has an 
initial downward velocity v of this velocity will, at the end of the 
time t, become 

Vss % = *«> + (iTsinfl)/ (2) 

at 

,\ s = vj + ig sin 6 . P, (3) 

and, for the unbalanced force acting, we have 

-F=m — = —(gsm$) = WsinO. (4) 

at 2 g 



94 APPLIED MECHANICS. 

If, on the other hand, the body's initial velocity is upward, 
and we denote this upward velocity by v ot we shall have the 
equations 

v = d ±^v Q - Or sin 0)/ (5) 

s = v t - \g sin . / 2 (6) 

F= -IVsinO. (7) 

Again, if the initial velocity is zero, equations (2) and (3) 
become 

v = f = (gsin6)t, (8) 

at 

s = \gsmB.t\ (9) 

From these we obtain, for this case, 



-1 



g sin# 
and, substituting this value of / in (8), we have 



(10) 



v = V2£(.r sin#), (11) 

or, if we let s sin $ = h = the vertical distance through which 
the body has fallen, we have 

v = ^2gh. (12) 

Hence, When a body, starting from rest, falls, under the 
action of gravity only, through a height h, the velocity acquired 
is V^2gh, whether the path be vertical or inclined. 

EXAMPLES. 

1. A body moves from the top to the bottom of a plane inclined 
to the horizon at 30 , under the action of gravity only : find the time 
required for the descent, and the velocity at the foot of the plane. 



MOTION ALONG A CURVED LINE. 



95 



2. In the right-angled triangle shown in the figure (Fig. 51 
AB = 10 feet, angle BAC = 30 : find the time a 
body would require, if acted on by gravity only, to fall 
from rest through each of the sides respectively, AB 
being vertical. 

3. Given inclination of plane to the horizon = 0, 
length of plane = /.- compare the time of falling down 
the plane with the time of falling down the vertical. 

4. A 100-pound weight rests, without friction, on the 
plane of example 3. What horizontal force is required 
to keep it from sliding down the plane. 

- 5. Suppose 5 pounds horizontal force to be applied 
(a) so as to oppose the descent, (b) so as to aid the descent : 
each case how long it will take the weight to descend from the 
the bottom plane. 




§ 80. Motion along a Curved Line under the Action of 
Gravity only. — We shall consider two questions in this 
regard : (a) the velocity at any point of the curve (b) the time 
of descent through any part of the curve. 

(a) Velocity at any point. Let us suppose the body to have 

started from rest at A, and to have 
reached the point P in time t, 
where AB = x (Fig. 52). Then, 
since the curved line AP may be 
considered as the limit of a broken 
line running from A to P, and as 
it has already been seen that the 
velocity acquired by falling through 
~ c a certain height depends only upon 
the height, and not upon the incli- 
nation of the path, we shall have for a curved line also 

V = \2gAB = \2gX, 

where v is the velocity at P. 




Fig. 52. 



96 APPLIED MECHANICS. 

(b) Time down a curve. Referring to the same figure, let t 
denote the time required to go from A to P, and At the time to 
go from P to P\ where PP' = As, and BB' = Ax; then, as we 
have seen that the velocity at P is \2gx, we shall have approx- 
imately for the space passed over in time A/, the equation 



or, passing to the limit, 



This equation gives 



or 



As — ^2gxAt, 



I = v^. (1) 



*= ds 



Shgx 



vMsy 



_ f ds _ r V \dxj (2) 

J ^2gX J 



dx 

2gX 



where, of course, the proper limits of integration must be 
used. 

If t denote the time from A to P, we have 

ds 



-/ 



2gX 



EXAMPLE. 



'c n 

Fig. 53. k. 



A body acted on by gravity only is constrained to 
move in the arc of a circle from A to C (Fig. 53), radius 
10 feet. Find the time of describing the arc (quadrant) 
and the velocity acquired by the body when it reaches 



SIMPLE CIRCULAR PENDULUM. 



97 



§81. Simple Circular Pendulum. — To find the time occu- 
pied in a vibration of a simple circu- ^c 
lar pendulum, we take D (Fig. 54) as 
origin, and DC as axis of x, and the 



axis of y at right angles to DC. Let 
AC = /and BD — //, we shall have 
for the time of a single oscillation 
from A to E 



r 



ds 




Now, from the equation of the circle AFDE, 

y 2 = 2/x — x 2 , 



we have 



*/y _ I — x 
dx y 

ds I I 



dx . y s/ 2 /x 



. / 



■/ 



Idx 



\J{2lx — X 2 ) \_2g{k — X)~\ \2gJ >J/lX — X 2 ^2/.:— X 



2/ p* 



dx 



or 



' y gj >lhx - x 2 \ 2// 



This can only be integrated approximately. 
• Y we obtain 



Expanding ( 1 

\ 2/y 



(-r= 



1 + -, + —*; + etc., 

4/ 32 / 2 



^/(■ + 5 + r?' + -)s=r 



98 APPLIED MECHANICS. 

The greatest value of x is h; and if h is so small that we may 
x_ 



omit — , we shall have as our approximate result 



If, however, the value of h as compared with / is too large 

to render it sufficiently accurate to omit — , but so small that 

4/ 

we can safely omit the higher powers of -, we shall have 

t = i/-^versm 1 I y 

V g { h 4/J Q sjkx - x 2 ) Q 

4 /7( . ~ x 2x . 1 rh . ~* 2x k n -"])* 

= l/-<versm 1 -versin — \hx — x 2 \\ 

V g \ h 4/L2 h J j o, 

or 

a nearer approximatioa 
The formula 

is the most used, and is more nearly correct, the smaller the 
value of h. 

EXAMPLES. 

1 . Find the length of the simple circular pendulum which is to beat 
seconds at a place where g = 32^-. 

Solution. 

-4, :•:■:<-:?= -oSf" *•»**• 



SIMPLE CYCLOID A L PENDULUM. 



99 



2. What is the time of vibration of a simple circular pendulum 5 
feet long? 

§82. Simple Cycloidal Pendulum. — The equation of the 

cycloid is 

~ z x 
y = atversin — h (2ax — x 2 )*, 
a 



dy — * 2a ~~ x 

dx V x 
ds _ (2aSS 
dx \ x J 

Hence we shall have, for the time of a single oscillation, 



a C dx 

7J slhx- 



X 2 



or 



(a\^ ( . l 2x) h .fa 



g 

This expression is independent of h> so that the time of vibra- 
tion is the same whether the arc be large or small. 

A body can be made to vibrate in a cycloidal arc by suspend- 
ing it by a flexible string between two cycloidal cheeks. This 
is shown from the fact that 
the evolute of the cycloid is 
another cycloid (Fig. 55). 

To prove this, we have, 
from the equation of the 
cycloid, 

y = a versin — f- (2ax — x 2 )*, 
a 



dy _ l 2a — x d s _ » 1 2a 

dx-\^T~' dx~-\^' 




d 2 y _ —a 



dx 2 x\sj 2a _ x 






IOO APPLIED MECHANICS. 



Hence the radius of curvature is 



p = 



d 2 y 



— 2(2a)*\2a — x; 



dx 2 

and since we have for the evolute the relation 

ds' = dp, 

where ds' is the elementary arc of the evolute, 

J^x = 2a 
dp; 
X — X 

and, observing that when x — 2a p = o, we have 

s' = p, 



.'. / = 2(2tf)^V / '2tf — X. 

If x l is the abscissa of the point of the evolute, 
x x — x + 



dv 
p-j = 4a — x, 

ds 



.'. s' = 2(2a)-\Xj — 2a; 

and, transforming co-ordinates to B by putting x i -|- 2a for x„ 

we obtain 

/ = 2(2^^2)5, 

,\ J 2 = Sax t , 

which is the equation of another cycloid just like the first. 

The motion along a vertical cycloid may also be obtained by 
letting a body move along a groove in the form of a cycloid 
acted on by gravity alone ; and in this case the time of descent 
of the body to the lowest point is precisely the same at what- 
ever point of the curve the body is placed. 

§ 83. Effect of Grade on the Tractive Force of a Rail- 
way Train. — -As a useful particular case of motion on an 
inclined plane, we have the case of a railroad train moving up 
or down a grade. It is necessary that a certain tractive force 



EFFECT OF GRADE ON TRACTIVE FORCE. 10 1 

be exerted in order to overcome the resistances, and keep 
the train moving at a uniform rate along a level track. If, 
on the other hand, the track is not on a level, and if we 
resolve the weight of the train into components at right angles 
to and along the plane of the track, we shall have in the latter 
component a force which must be added to the tractive force 
above referred to when we wish to know the tractive force re- 
quired to carry it up grade, and must be subtracted when we 
wish to know the tractive force required to carry it down grade. 
The result of this subtraction may give, if the grade is suffi- 
ciently steep and the speed sufficiently slow, a negative quan- 
tity ; and in that case we must apply the brakes, instead of 
using steam, unless we wish the speed of the train to increase, 

EXAMPLES. 

i. A railroad train weighing 60000 lbs., and running at 50 miles per 
hour, requires a tractive force of 618 lbs. on a level ; what is the tractive 
force necessary when it is to ascend a grade of 50 feet per mile? What 
when it is to descend? Also what is the amount of work per minute 

in each case? 

Solution. 

The resolution of the weight will give (Fig. 50, § y? •, for the com- 
ponent along the plane, 



Hence 



( 6oooo )i& = 568.2 nearly. 



Tractive force for a level = 618.0, 
Tractive force for ascent = 1 186.2, 
Tractive force for descent = 49.8. 

To ascertain the work done per minute in each case, we have — 
(a) For a level track, 6l8 x 5 ^ x 528 ° = 2719200 foot-lbs. 

(d) Up grade, 2719200 -f 6ooo ° x 6o 5 ° x 5 ° = 5219200 foot-lbs. 
(c) Down grade, 2719200 — 6ooo ° * 5 ° x 5 ° = 219200 foot-lbs. 



102 



APPLIED MECHANICS. 



2. Suppose the tractive force required for each 2000 lbs. of weight 
of train to be, on a level track, for velocities of — 

5.0 miles per hour, 10.0 20.0 30.0 40.0 50.0 60 

6.1 lbs., 6.6 8.3 11. 2 15.3 20.6 27; 
find the tractive force required to carry the train of example 1 — 

(a) Up an incline of 50 feet per mile at 30 miles per hour. 

(b) Down an incline of 50 feet per mile, at 30 miles per hour. 

(c) Down an incline of 10 feet per mile at 20 miles per hour. 

(//) What must be the incline down which the train must run to 
require no tractive force at 40 miles per hour? 

3. If in the first example the tractive force remains 618 lbs. while 
the train is going down grade, what will be its velocity at the end of one 
minute, the grade being 10 feet per mile? 



§84. Harmonic Motion. — If we imagine a body to be 
moving in a circle at a uniform rate (Fig. 56), and a second 

body to oscillate back and forth in 
the diameter AB, both starting 
from B, and 

if when the /"~T~~\ 

first body is / \ 

at C the other 
is directly un- 
der it at G, 
etc., then is 
the second 
body said to 
move in harmonic motion. 

A practical case of this kind of mo- 
tion is the motion of a slotted cross-head 
of an engine, as shown in the figure 
(Fig. 57) ; the crank moving at a uni- 
form rate. In the case of the ordinary 
crank, and connecting-rod connecting 
the drive-wheel shaft of a stationary engine with the piston-rod, 




Fig. 56. 




HARMONIC MOTION. IO3 

we have in the motion of the piston only an approximation to 
harmonic motion. We will proceed to determine the law of the 
force acting upon, and the velocity of, a body which is con- 
strained to move in harmonic motion. Let the body itself and 
the corresponding revolving body be supposed to start from 
B (Fig. 56), the latter revolving in left-handed rotation with an 
angular velocity a, and let the time taken by the former in 
reaching G be /: then will the angle BOC = at; and we shall 
have, if s denote the space passed over by the body that moves 
with harmonic motion, 

s = BG = OB — OCcosat, 
or, if 

r = OB = OC, 

s = r — r cos at, (l) 

the velocity at the end of the time / will be 

ds . , x 

V = — = arsina/, (2) 

and the acceleration at the end of time t will be 

/= -—= a 2 r COS at. (3) 

Hence the force acting upon the body at that instant, in the 
direction of its motion, is 

d 2 s 
F— m — = ma 2 rcosat = ma 2 (OG). (4) 

The force, therefore, varies directly as the distance of the body 
from the centre of its path. It is zero when the body is at the 



104 APPLIED MECHANICS. 

centre of its path, and greatest when it is at the ends of its 

travel, as its value is then 

W 
ma 2 r == — a*r; 

g 

this being the same in amount as the centrifugal force of the 
revolving body, provided this latter have the same weight as the 
oscillating body. On the other hand, the velocity is greatest 

when at ..== - (i.e., at mid-stroke) ; and its value is then 
2 

v = ar, 
this being also the velocity of the crank-pin at mid-stroke. 

EXAMPLE. 

Given that the reciprocating parts of an engine weigh ioooo lbs., 
the length of crank being i foot, the crank making 60 revolutions per 
minute ; find the force required to make the cross-head follow the crank, 
(1) when the crank stands at 30 to the line of dead points, (2) when 
at 6o°, (3) when at the dead point. 

§ 85. Work under Oblique Force. — If the force act in 
any other direction than that of the motion, we must resolve it 
into two components, the component in the direction of the 
motion being the only one that does work. Thus if the force 
F is variable, and 6 equals the angle it makes with the direction 
of the motion, we shall have as our expression for the work 
done 

/ F 'cos Ods. 

Thus if a constant force of 100 lbs. act upon a body in a direc- 
tion making an angle of 30 with the line of motion, then wil( 
the work done by the force during the time in which it moves 
through a distance of 10 feet be 

(100) (0.86603) (10) = 866 foot-lbs. 




ROTATION OF RIGID BODIES. 105 

§ 86. Rotation of Rigid Bodies. — Suppose a rigid body 
(Fig. 58) to revolve about an axis perpendicular to the plane of 
the paper, and passing through O ; 
imagine a particle whose weight is 
w to be situated at a perpendicular 
distance OA == r from the axis of 
rotation, and let the angular accel- 
eration be a : let it now be required 
to find, the moment of the force or 
forces required to impart this ac- 
celeration ; for we know that, if 

the axis of rotation pass through the centre of gravity of the 
body, the motion can be imparted only by a statical couple ; 
whereas if it do not pass through the centre of gravity, the 
motion can be imparted by a single force. 

We shall have, for the particle situated at A, 

Weight — w. . 

Angular acceleration = a. 

Linear acceleration == ar. 

Force required to impart this acceleration to this particle 

w 

= — ar. 

S 

Moment of this force about the axis == — ar 2 . 

g 

Hence the moment of the force or forces required to impart 
to the entire body in a unit of time a rotation about the axis 
through O, with an angular velocity a, is 

2,—ar 2 = -^wr 2 = — , 
g g g 

where / is used as a symbol to denote the limit of ^wr 2 , and is 
called the Moment of Inertia of the body about the axis through O. 



106 APPLIED MECHANICS. 

\%j. Angular Momentum. — This quantity, — , which ex- 

g 
presses the moment of the force or forces required to impart to 
the body the angular acceleration a about the axis in question 
is also called the Angular Momentum of the body when rotat- 
ing with the angular velocity a about the given axis. 

§ 88. Actual Energy of a Rotating Body. — If it be re- 
quired to find the actual energy of the body when rotating 
with the angular velocity co, we have, for the actual energy of 
the particle at A, 

w (a>r) 2 a? 

— - — — = — wr 2 t 

g 2 2g 

and for that of the entire body 

2g *g 

This is the amount of mechanical work which would have to be 
done to bring the body from a state of rest to the velocity <o, or 
the total amount of work which the body could do in virtue 
of its velocity against any resistance tending to stop its 
rotation. 

§ 89. Moment of Inertia. — The term "moment of inertia ! ' 
originated in a wrong conception of the properties of matter. 
The term has, however, been retained as a very convenient one, 
although the conceptions under which it originated have long 
ago vanished. The meaning of the term as at present used, in 
relation to a solid body, is as follows : — 

The moment of inertia of a body about a given axis is the 
limit of the sum of the products of the weight of each of the ele- 
mentary particles that make up the body, by the squares of their 
distances from the given axis. 

Thus, if w If w 2 , w v etc., are the weights of the particles 
which are situated at distances r lf r 2 , r v etc., respectively from 



MOMENT OF INERTIA OF A PLANE SURFACE. I0y 

the axis, the moment of inertia of the body about the given 

axis is 

/ = limit of "Swr 2 . 

§ 90. Radius of Gyration. — The radius of gyration of a 
body with respect to an axis is the perpendicular distance from 
the axis to that point at which, if the whole mass of the body 
were concentrated, the angular momentum, and hence the mo- 
ment of inertia, of the body, would remain the same as they are 
in the body itself. 

If p is the radius of gyration, the moment of inertia would 
be, when the mass is concentrated, 

p 2 2w ; 
hence we must have 

p 2 2,w — %wr 2 = I, 
whence 

Zwr 2 I 



P 2 = 



^w W* 



where W = entire weight of the body. 

§91. Moment of Inertia of a Plane Surface. — The term 

"moment of inertia," when applied to a plane figure, must, of 
course, be defined a little differently, as a plane surface has no 
weight ; but, inasmuch as the quantity to which that name is 
given is necessary for the solution of a great many questions. 

The moment of inertia of a plane surface about an axis, either 
in or not in the plane, is the limit of the sum of the products of 
the elementary areas into which the surface may be conceived to 
be divided, by the squares of their distances frovi the axis in 
question. 

In a similar way, for the radius of gyration p of a plane 
figure whose area is A, we have 



108 APPLIED MECHANICS. 

From this definition it will be evident, that, if the surface be 
referred to a pair of axes in its own plane, the moment of iner- 
tia of the surface about O Y will be 

I=ffx 2 dxdy, (i) 

and the moment of inertia of the surface about OX will be 

J^fSfdxdy. (2) 

The moment of inertia of the surface about an axis passing 
through the origin, and perpendicular to the plane XO Y, will be 

ffr*dxdy, (3) 

where r = distance from O to the point (x f y) ; hence r 2 — x 2 -f- 
y 2 , and the moment of inertia becomes 

ff (*2 + y^dxdy = Sfx-dxdy + fffdxdy = / + /. (4) 

This is called the "polar moment of inertia." If polar co-ordi- 
nates be used, this last becomes 

ffp 2 (pd P dO) = ffpidpdO. (5) 

All these quantities are quantities that will arise in the discus- 
sion of stresses, and the letters /and./ are very commonly used 
to denote respectively 

ffx 2 dxdy and ffy 2 dxdy. 

Another quantity that occurs also, and which will be repre- 
sented by K, is 

ffxydxdy ; (6) 

and this is called the moment of deviation. 



EXAMPLES OF MOMENTS OF INERTIA. 



109 



EXAMPLES. 



The following examples will illustrate the mode of finding the 
moment of inertia : — .x 

1. Find the moment of inertia of the rectangle 
ABCD about OY (Fig. 59). 



Solution. 

h 



/= /u^ = v/ 



dx 



btf_ 
12 



A B 



C D 



Fig. 



2. Find the moment of inertia of the entire circle (radius r) about 
the diameter OY (Fig. 60). 




Fig. 60. 



Solution. 



/ = / / x'dxdy = 2 / xWr 2 - 

t/_ r ti_v^z^ J t/_ r 



xdx 



nr* _ 7td i 
4 " 64 ' 



3. Find the moment of inertia of the circular ring (outside radius r, 
inside radius r x ) about OY (Fig. 61). 



Solution, 
irr* irrS ir{r* — rj) w(d* — dj) 



64 



4. Find the moment of inertia of an ellipse 
(semi-axes a and b) about the minor axis OY. 




Fig. 61. 



no 



APPLIED MECHANICS. 



Solution. 



x 2 - y 2 
Equation of ellipse is — + tz = i 



ly 



/:/; 



x^dxdy 



ao _ a a\ o / 



na*b 



On the other hand, I x — 



irab 3 



§ 92. Moments of Inertia of Plane Figures about Parallel 
Axes. 

Proposition. — The moment of inertia of a plane figure 
about an axis not passing through its centre of gravity is equal 
to its moment of inertia about a parallel axis pass i?ig through its 
centre of gravity increased by the product obtained by multiply- 
ing the area by the square of the distance between the two axes. 

Proof. — Let A B CD 
(Fig. 62) be the surface ; let 
O Y be the axis not passing 
through the centre of grav- 
ity ; let P be an elementary 
area AxAy, whose co-ordi- 
nates are OP = x and RP 
= y; and let 00 \ = a = a 
constant — distance be- 
tween the axes. 
Let 0,R = x T = abscissa of P with reference to the axis 
passing through the centre of gravity, 

x — a "+ x x 
x 2 = x* 4 20.*! + a 2 
,\ x 2 AxAy = x s *Ax Ay 4 2axAx&y 4 a 2 &xAy. 




POLAR MOMENT OF INERTIA OF PLANE FIGURES. Ill 



Hence, summing, and passing to the limit, we have 

ffx 2 dxdy — ffxfdxdy + 2affx l dxdy -f a 2 ffdxdy j (i) 

but if we were seeking the abscissa of the centre of gravity 
when the surface is referred to YfiY^ and if this abscissa be 
denoted by x oy we should have 

_ ffxjdxdy 

x °-'T/dx^r ; 

and, since x Q = o, .'. ffx^dxdy = o ; hence, substituting this 
value in (i), we obtain 

ffx 2 dxdy = ffx t 2 dxdy -f- a 2 ffdxdy. (2) 

If, now, we call the moment of inertia about OY, I, that 
about 1 Y 1} /„ and let the area = A = ffdxdy, we shall have 

/ = / f + a* A. (3) 

Q. E. D. 

§ 93. Polar Moment of Inertia of Plane Figures. — The 

moment of inertia of a plane 
figure about an axis perpen- 
dicular to the plane is equal 
to the sum of its moments 
of inertia about any pair of D t 
rectangular axes in its plane 
passing through the foot of 
the perpendicular. 

Proof. — Let BCD (Fig. 
63) be the surface, and P an 
elementary area, and let 
OA = x y AP — y, OP — r 
the surface about OZ will be 




Fig. 63. 

then the moment of inertia of 



ffr*dxdy =ff(x* +f)dxdy =ffx 2 dxdy + f ffdxdy = / -f /. 
Q.E.D. . 



112 APPLIED MECHANICS. 

Hence follows, also, that the sum of the moments of inertia 
of a plane surface relatively to a pair of rectangular axes in its 
own plane is isotropic ; i.e., the same as for any other pair of 
rectangular axes meeting at the same point, and lying in its 
plane. 

EXAMPLES. 

i. To find the moment of inertia of the rectangle (Fig. 59) about 
an axis through its centre perpendicular to the plane of the rectangle. 

Solution. 
Moment of inertia about YY = — , 

T2 

Moment of inertia about an axis through its 



hence 



centre and perpendicular to YY = — ; 

12 



Polar moment of inertia = 1 = — (h 2 -f £ 2 ). 

12 12 12 



2. To find the moment of inertia of a circle about an axis through 
its centre and perpendicular to its plane (Fig. 60). 

Solution. 

7rr* 
Moment of inertia about OY = — , 

4 

Moment of inertia about OX =; — ; 



hence 



„ , r . -n-r 4 -rrr* -n-r 4 

Polar moment of inertia = — -\ = — . 

442 



3. To find the moment of inertia of an ellipse about an axis passing 
through its centre and perpendicular to its plane. 



MOMENTS OF INERTIA ABOV 2* DIFFERENT AXES. il$ 



Solution. 
From example 4, § 91, we have 

j~ — 



/,= 



iratb 



Polar moment of inertia = {a 2 -f- £*). 

4 



§ 94. Moments of Inertia of Plane Figures about Different 
Axes compared. — Given the surface KLM (Fig. 64), suppose 
we have already determined the quantities 

/ = ffx 2 dxdy, J = ffy 2 dxdy, K = ffxydxdy, 

it is required to determine, in terms of them, the quantities 

** = ffx'dx.dy,, / r = ffy 1 2 dx l dy 1J K x = ffx^dx.dy, ; 



the angles Jf<9Fand X x OY t being both 
right angles, and YOY T = a. 

We shall have, from the ordinary 
equations for the transformation of co- 
ordinates, to be found in any analytic 
geometry, the equations 



x\ = x cos a -f y sin a, 

y z = y cos a — X sin a, 
^ x 2 = x 2 cos 2 a -f- j 2 sin 2 a -f 2jrv cos a sin a, 
y* = x 2 sin 2 a -f- y 2 cos 2 a — 24^ cos a sin a, 
jv i j i = xy(cos 2 a — sin 2 a) — (x 2 — y 2 ) cos a sin u. 




Fig. 64. 



U4 . APPIIED MECHANICS. 



Hence 



/, = ffx 2 dxjy x = limit of %x?AA 

= cos 2 a limit of %x 2 AA + sin 2 a limit of %y*&A 4- 

2 cos a sin a limit of %xyAA 
= (cos 2 a) ffx 2 dxdy + (sin 2 a) ffy 2 dxdy ■+■ 

2 (cos a sin a) ffxydxdy. 
J x = ffy 2 dx,dy T = limit of S^A^ 

= (sin 2 a) limit of %x 2 AA -f (cos 2 a) limit of 2/Ail — 

2 (cos a sin a) limit of ^xyAA 
= (sin 2 a) ffx 2 dxdy + (cos 2 a)ffy 2 dxdy — 

2 (cos a sin a) ffxydxdy. 
K t — ffx l y 1 dx 1 dy I = limit of %x I y 1 AA 

= (cos 2 a — sin 2 a) limit of 2.#yA^ — (cos a sin a) ^limit of 

2.* 2 A^ — limit of 2y 2 AA \ 
= (cos 2 a — sin 2 a) ffxydxdy — (cos a sin a) \ffx 2 dxdy — 

ffy 2 dxdy\. 

Or, introducing the letters /, /, and K, we have 

I x = I cos 2 a. -\- J sin 2 a + 2K cos a sin a, (1) 

J x = /sin 2 a + ycos 2 a — 2 A' cos a sin a, (2) 

.^ = (/— I) cos a sin a + ^(cos 2 a — sin 2 a). (3) 

The equations (1), (2), and (3) furnish the solution of the 
problem. 

§95. Principal Moments of Inertia in a Plane. — In every 
plane figure, a given point being assumed as origin, there is at 
least one pair of rectangular axes, about one of which the moment 
of inertia is a maximum, and a minimum about the other ; these 
moments of inertia being called principal moments of inertia, 
and the axes about which they are taken being called principal 
axes of inertia 



AXES OF SYMMETRY OF PLANE FIGURES. 115 



Proof. — In order that /„ equation (1), § 94, may be a maxi- 
mum or a minimum, we must have, as will be seen by differen- 
tiating its value, and putting the first differential co-efficient 
equal to zero, 

— 2/cos a sin a + 2J cos a sin a + 2iT(cos 2 a — sin 2 a) = o 

/. ^T(cos 2 a — sin 2 a) — (/ — /) cos a sin a = o ( 1) 

cos a sin a K . . „ ' 2K , x 

tan 2 a = . (2) 



cos 2 a — sin 2 a I — J I — J 

Hence, for the value of a given by (2), we have /, a maximum 
or a minimum ; and as there are two values of 2a corresponding 
to the same value of tan 2a, and as these two values differ by 
180 , the values of a will differ by 90 , one corresponding to a 
maximum and the other to a minimum. 

Moreover, when the value of a is so chosen, we have 

K* = o, 

as is proved by equation (1). Indeed, we might say that the 
condition for determining the principal axes of inertia is 

K x = 0. 

§ 96. Axes of Symmetry of Plane Figures. — An axis 
which divides the figure symmetrically is always a principal 
axis. 

Proof. — Let us assume that the y axis divides the surface 
symmetrically ; then we shall have, with reference to this axis, 

K = J J^xydydx = | J X -ydy [ _ = o. 

And, since K is zero, the axis of y is one principal axis, and of 
course the axis of x is the other. The same method of reason- 
ing would show K = o if the x axis were the axis of symmetry. 



Il6 APPLIED MECHANICS. 

Hence, whenever a plane figure has an axis of symmetry, 
this axis is one of the principal axes, and the other is at 
right angles to it. Thus, for a rectangle, when the axis is to 
pass through its centre of gravity, the principal axes are par- 
allel to the sides respectively, the moment of inertia being 
greatest about the shortest axis, and least about the longest. 
Thus in an ellipse the minor axis is the axis of maximum, 
and the major that of minimum, moment of inertia, etc. On 
the other hand, in a circle, or in a square, since the maximum 
and minimum are equal, it follows that the moments of inertia 
about all axes passing through the centre are the same. 

§ 97. Conditions for Equal Values of Moment of In- 
ertia. — When the moments of inertia of a plane figure about 
three different axes passing through the same point are the 
same, the moments of inertia about all axes passing through 
this point are the same. 

Proof. — Let I be the moment of inertia about O Y, I t 
about O Y lf I 2 about O Y 2 , and let 

YOY, = a, YOY 2 = ft 
and let 

/, = h = I. 

Then, from equation (1), § 94, we have 

I — I cos 2 a -J- J sin 2 a. -\- 2 K cos a sin a, 

/= / cos 2 /? -f /sin 2 /? -h 2 ^Tcos/3sinft 



Hence 



Hence 



(/ — y)sin 2 a = 2 K cos a sin a, (1) 

(/-/)sin 2 /?= 2 ^cos/3sin/3. (2) 



(/-/)tana = 2K, ■ (3) 

(i->j)uuip=2x: (4) 

And, since tan a is not equal to tan /?, we must have 
/ — J = o and K = o. 

Hence, since K — o and f =J, we shall have, from eqja- 



MOMENTS OF INERTIA ABOUT PARALLEL AXES. l\J 

tion (i), §94, for the moment of inertia /' about an axis,, 
making any angle 6 with Y, 

I' = /cos 2 6 + /sin 2 0-\-o = I. (5) 

Hence all the moments of inertia are equal. 

§ 98. Components of Moments of Inertia of Solid 
Bodies. — Refer the body to three rectangular axes, OX, OY, 
and OZ; and let I x , I y , and I z represent its moment of inertia 
about each axis respectively. Then, if r denote the distance of 
any particle from OZ> we shall have 

I z = limit of ^wr 2 ; 
but 

r 2 = x 2 -f y 2 

. * . I z = limit of %w(x 2 -f y 2 ) = limit of *%wx 2 -\- limit of Ikwy 2 . (1) 

In the same way we have 

Ijc = limit of %wy 2 + limit of %wz 2 , (2) 

I y = limit of ~%wx 2 4- limit oCSwz 2 . (3) 

§ 99. Moments of Inertia of Solids around Parallel 
Axes. — The moment of inertia of a solid body about an axis 
not passing through its centre of gravity is equal to its moment 
of inertia about a parallel axis passing through the centre of 
gravity, increased by the product of the entire weight of the 
body by the square of the distance between the two axes. 

Proof. — Refer the body to a system of three rectangular 
axes, OX, OY, and OZ, of which OZ is the one about which 
the moment of inertia is taken. Let the co-ordinates of the 
centre of gravity of the body with reference to these axes be 
(*o> 7o» #o)- Through the centre of gravity of the body draw a 
system of rectangular axes, parallel respectively to OX, OY, and 
OZ. Then we shall have for the co-ordinates of any point 

X — X ""|— Xjj 

y = y Q + y„ 

z = z Q + z lt 



u8 



APPLIED MECHANICS. 



Hence 

I z = limit of %w (x 2 -h y 2 ) = limit of 3zeA* 2 4- limit of 2ze/y 2 
= limit of ^,w(x 4- ^ t ) 2 + limit of 2w(jy + y t ) 2 
= jc 2 limit of Ioju -f 7o 2 limit of 3ze> + 2.r limit of %wx t 

-f- 2y Q limit of %wy l 4- limit of ^wXj 2 + limit of ^ze^ 2 
= {Xo + y Q 2 )W + 2x Q limit of %wx l -f 2jy limit of 2ze/)\ 

4- limit of ^wrf 
= r 2 W + I z f 4- 2^ limit of Sowi 4- 2jy limit of So^x. 

But, since 1 is the centre of gravity, 

.*. Sawi = o and So^ = o. 
Hence 

J z = // 4- Wr*, 

which proves the proposition. 

§ ioo. Examples of Moments of Inertia. 

i . To find the moment of inertia of a sphere whose radius is r and 
weight per unit of volume w, about the axis OZ drawn through its centre. 

Solution. 

Divide the sphere into thin slices (Fig. 65) by planes drawn perpen- 
dicular to OZ. Let the distance 
of the slice shown in the figure, 
above O be z, and its thickness dz : 




then will- its radius be sir 2 — z 2 ; 
and we can readily see, from ex- 
ample 2, § 93, that its moment of 



inertia about OZ will be 
ww(r 2 — z 2 )'- 



Fig. 65. 



= W - f l 
2 J-r 



dz. 



Hence the moment of inertia 
of the entire sphere about OZ will 
be 

{f — z 2 ) 2 dz, 



EXAMPLES OF MOMENTS OF INERTIA. 



II 9 



which easily reduces to 



_8_ 
15 



Wirr*. 



2. To find the moment of inertia of an ellipsoid (semi-axes a, b, c) 
about OZ (Fig. 66). 

Solution. — The equa- 
tion of the ellipsoid is 



f 



a 2 + b 



1. 



c* 



Divide it into thin slices 
perpendicular to OZ, and 
let the slice shown in the 
figure be at a distance z 
from O. Then will this 
slice be elliptical, and its 
semi-axes will be 




Fig. 66. 



and 



- sic 2 - z 2 ; 



and from example 3, § 93, we readily obtain, for its moment of inertia 
about OZ, 

TB ( ^- 22) ]i^ 2 - 22) + ^ 2 - 22) l^ 

= wwab(* + *) - ( _ yA 

4 c* 

Hence, for the moment of inertia of the ellipsoid about OZ, we 



have 



/« = 



wirabid 2 4- b 2 ) C c , x , 4 

^— — — L I (c 2 - z 2 ) 2 dz = -±- WTvabc(a 2 + b 2 ). 

A c J-c 15 



3. Find the moment of inertia of a right circular cylinder, length a, 
radius r, about its axis. 

Ans. 



120 APPLIED MECHANICS. 

4. Find the moment of inertia of the same about an axis perpen- 
dicular to, .and bisecting its axis. * wirar 2 1 2 •, a 2 \ 

4 V 3/ 

5. Find the moment of inertia of an elliptic right cylinder, length 

2c, transverse semi-axes a and b, about its longitudinal axis. 

Ans. {a 2 + b 2 ). 

2 

6. Find the moment of inertia of the same about its transverse 
axis 2b. 



Ans. 2WTzabc 



(M> 




7. Find the moment of inertia of a rectangular prism, sides 2a, 2b, 
2c, about central axis 2c. Ans. %wabc(a 2 + b 2 ). 

§ 101. Centre of Percussion. — Suppose we have a body 
revolving, with an angular velocity a, about an axis perpendicu- 
lar to the plane of the paper, and 
passing through O. Join O with 
the centre of gravity, G, and take OG 
as axis of x ; the axis of y passing 
through O, and lying in the plane of 
the paper. If, with a radius OA == r, 
we describe an arc CA (Fig. 67), all 
FlG ' 67 ' particles situated in this arc have a 

linear velocity ar. The force which would impart this velocity 
to any one of them, as that at A, in a unit of time, is 

w 
-ar, 

g 
and this may be resolved into two, 

w •, w 

—ax and — ay, 

g g 

respectively perpendicular and parallel to OG. The moment of 

this force about the axis is 

w , 
—ar 2 ; 

g 
hence the total moment of the forces which would impart to 



CENTRE OF PERCUSSION. 121 



the body in a unit of time the angular velocity a, is, as has been 
shown already, 

— = —%wr 2 . 
g g 

The resultant of the forces acting on the body is 

-%wx, 

g 

since, the centre of gravity being on OB, it follows that 
%wy = o ; and hence 

-%wy = o. 
g 

Hence the perpendicular distance from O to the line of direc- 
tion of the resultant force is measured along OG, and is 

£ 
and the point of application of the resultant force may be con- 
ceived to be at a point on OG at a distance / from O ; and this 
point of application of the resultant of the forces which pro- 
duce the rotation is called the Centre of Percussion. 

li p =z radius of gyration' about the axis through O, and if 
x = distance from O to the centre of gravity, we have 

Hence 



^wx x Q %w x\W J X Q 



or, in words, — 

The radius of gyration is a mean proportional between the 
distance 1, and the distance x , between the axis of oscillation and 
the ce7itre of gravity. 

The centre of percussion with respect to a given axis of 
oscillation O has been defined as the point of application of the 



/ 



122 APPLIED MECHANICS. 

resultant of the forces which cause the body to rotate around ike 
point O. 

Another definition often given is, that it is the point at which, 
if a force be applied, there will be no shock on the axis of oscilla~ 
tion; and these two definitions are equivalent to each other. 

Let the particles of the body under consideration be con- 
ceived, for the sake of simplicity, to be distributed along a single 
line AB, and suppose a force F applied at D 
(Fig. 68). Conceive two equal and opposite 
forces, each equal to F, applied at C, the cen- 
tre of gravity of the body. 
\|/ c Then these three forces are equivalent to 

a single force F applied at the centre of grav- 
ity C, which produces translation of the whole 
body ; and, secondly, a couple whose moment 
is F{CD) y whose effect is to produce rotation 
Fig. 68. around an axis passing through the centre of 

gravity C. Under this condition of things it is evident that the 
centre of gravity C will have imparted to it in a unit of time a 

forward velocity equal to — , where M is the entire mass of the 

body ; the point D will have imparted to it a greater forward 
velocity ; while those points on the upper side of C will have 
imparted to them a less and less velocity as they recede from 
C, until, if the rod is sufficiently long, the particle at A will 
acquire a backward velocity. 

Hence there must be some point which for the instant in 
question is at rest; i.e., where the velocity due to rotation is just 
equal and opposite to that due to the translation, or about 
which, for the instant, the body is rotating : and if this point 
were fixed by a pivot, there would be no stress on the pivot 
caused by the force applied at D. 

An axis through this point is called the Instantaneous 
Axis. 



IMPACT OR COLLISION. 1 23 

§ 102. Interchangeability of the Centre of Percussion 
and Axis of Oscillation. — If we take, as axis of oscillation, a 
line perpendicular to the plane of the paper, and passing 
through D, then will O be the new centre of percussion. 

Proof. — We have seen (§ 101) that 

'-£ 

where / = OD, x = OC, and p = radius of gyration about an 
axis through O perpendicular to the plane of the paper. 

Moreover, if p represent the radius of gyration about an 
axis through C perpendicular to the plane of the paper, we shall 
have 

P 2 = Po 2 + Xo 2 

Xo 

.-. /-*. = £= CD. 

Xo 

Now if D is taken as axis of oscillation, we shall have for the 
distance / x to the corresponding centre of percussion, 

/ r= p * 2 = ^ 
1 CD I - xo 

where p x = radius of gyration about the axis of oscillation 
through D. 

' I - pI * - p ° 2 + CD2 - p ° 2 4- CD -x 4- (f-x) - I 
' l ~ CD~ CD ~ CD + D ~ X ° + ( o) " 

Hence the new centre of percussion is at O. Q. E. D. 

§ 103. Impact or Collision. — Impact or collision is a 
pressure of inappreciably short duration between two bodies. 

The direction of the force of impact is along the straight line 
drawn normal to the surfaces of the colliding bodies at their 
point of contact, and we may call this line the line of impact. 



124 APPLIED MECHANICS. 

The action that occurs in the case of collision may be de- 
scribed as follows : at first the bodies undergo compression ; 
the mutual pressure between them constantly increasing, until, 
when it has reached its maximum, the elasticity of the mate- 
rials begins to overpower the compressive force, and restore 
the bodies wholly or partially to their original shape and xlimen- 
sions. 

Central impact occurs when the line joining the centres of 
gravity of the bodies coincides with the line of impact. 

Eccentric impact occurs when these lines do not coincide. 

Direct impact occurs when the line along which the relative 
motion of the bodies takes place, coincides with the line of 
impact. 

Oblique impact occurs when these lines do not coincide. 

CENTRAL IMPACT. 

§ 104. Equality of Action and Re-action. — One funda- 
mental principle that holds in all cases of central impact is the 
equality of action and re-action ; in other words, we must have, 
that, at every instant of the time during which the impact is 
taking place, the pressure that one body exerts upon the other 
is equal and opposite to that exerted by the second upon the 
first. 

The direct consequence of this principle is, that the algebraic 
sum of the momenta of the two bodies before impact remains 
unaltered by the impact, and hence that this sum is just the 
same at every instant of, and after, the impact. 

If we let 

tn„ m„ be the respective masses, 
e„ c 2 , their respective velocities before impact, 

v„ v„ their respective velocities after impact, 

i/ t v"> their respective velocities at any given instant during 
the time while impact is taking place, 



CO-EFFICIENT OF RESTITUTION. 12$ 

then we must have the following two equations true ; viz., — 

m^j -f- m 2 V2 = ^/i -f- m 2 c 2, (i) 

m x v' + m 2 v" = »*i*j + w 2 f 2 . (2) 

§105. Velocity at Time of Greatest Compression. — At 

the instant when the compression is greatest — i.e., at the 
instant when the elasticity of the bodies begins to overcome 
the deformation due to the impact, and to tend to restore them 
to their original forms — the values of v' and v" must be equal 
to each other; in other words, the colliding bodies must be 
moving with a common velocity 

*'= j = tT. (1) 

To determine this velocity, we have, from equation (2), § 104, 
combined with (1), 

v _ m ^ + m 2 c 2 ^ ^ 

tn l 4- m 2 

§ 106. Co-efficient of Restitution.- — In order to determine 
the values v u v 2> of the velocities after impact, we need two 
equations, and hence two conditions. One of them is fur- 
nished by equation (1), § 104. The second depends upon the 
nature of the material of the colliding bodies, and we may dis- 
tinguish three cases : — 

i°. Inelastic Impact. — In this case the velocity lost up to 
the time of greatest compression is not regained at all, and 
the velocity after impact is the common velocity v at the instant 
of greatest compression. In this case the whole of the work 
used up in compressing the bodies is lost, as none of it is 
restored by the elasticity of the material. 

2°. Elastic Impact. — In this case the velocity regained 
after the greatest compression, is equal and opposite to that 
lost up to the time of greatest compression ; therefore 

v — V t = € v — v. (1) v 2 — V = v — c 2 . (2) 



126 APPLIED MECHANICS. 

We may also define this case as that in which the work lost 
in compressing the bodies is entirely restored by the elasticity 
of the material, so that 

1 = 1 . (3) 

2 2 2 2 

Either condition will lead to the same result. 

3°. Imperfectly Elastic Impact. — In this case a part only 
of the velocity lost up to the time of greatest compression is 
regained after that time. 

If, when the two bodies are of the same material, we call e 
the co-efficient of restitution, then we shall so define it that 

v — v t v 2 — v 

C % — V ~~ V — c 2 " 

or, in words, the co-efficient of restitution is the ratio of the 
velocity regained after compression to that lost previous to 
that time. 

In this case only a part of the work done in producing the 
compression is regained, hence there is loss of energy. Its 
amount will be determined later. 

Strictly speaking, all bodies belong to the third class ; the 
value of e being always a proper fraction, and never reaching 
unity, the value corresponding to perfect elasticity ; nor zero, 
the value corresponding to entire lack of elasticity. 

§ 107. Inelastic Impact. — In this case the velocity after 
impact is the common velocity at the time of greatest com- 
pression ; hence 

v = z/j = v 2 , (1) 

m . m v= m x c x + m 2 c^ ^ 

m t -f- m 2 

And for the loss of energy due to impact we have 

-i-i- + -2-2- — (m x -f m 2 ) — , 
2 2 2 



ELASTIC IMPACT. 12J 



which, on substituting the value of v, reduces to 



m,m. 



2{m 1 + m 2 ) 



('i - ^) a - (3) 



§ 1 08. Elastic Impact. — In this case we have, of course, 
the condition, equation (1), § 104, 

m x v x + m 2 v 2 = m I c I + m 2 c 2i 

and, for second equation, we may use equation (3), § 106 ; viz., 

m x v x 2 m 2 v 2 2 _ m^c? m^f 
2222 

Combining these two equations, we shall obtain 

Vl m ,, _ *»*(,<. - <*) , (I) 

«, + m 2 

V2 = ,, + »»■('■ - *> . (i ) 

w x 4- *# 2 

We can obtain the same result without having to solve an 
equation of the second degree, by using instead the equations 
(1) and (2) of § 106, together with (1) of § 104; i.e.,— 





m x v x + ^2^2 = m x c x + m 2 c t ; 




v — v x = c x — V, 


or 






v 2 — V = V — c 2 , 


and (§ 105) 






_ m x c x + m 2 c 2 



Ml + ^2 

As the result of combining these equations, and eliminating 
v, we should obtain equations (1) and (2), as above, for the values 
of v x and v 2 . In this case the energy lost by the collision is 
zero. 



128 APPLIED MECHANICS. 

§ 109. Special Cases of Inelastic Impact. — (a) Let the 
mass m 2 be at rest. Then c 2 = o, 

(1) 



m l -\- m 



m^ntz c x 



.' . Loss of energy = Wl '" 2 ^. (2) 

m l -\- m 2 2 

(b) Let m 2 be at rest, and let m 2 = 00 ; i.e., let the mass m t 
strike against another which is at rest, and whose mass is in- 
finite. We have 

m 2 = 00, c 2 = o, 



v = 



Loss of energy = 



v - °> 


(3) 


0*i ^i 2 m^c? 


(4) 


m, x 2 2 



;;/, 



or the moving body is reduced to rest by the collision, and all 
its energy is expended in compression. 

(c) Let m l c I = —m 2 c 2 ; i.e., let the two bodies move towards 
each other with equal momenta : 

... y = m * c * + m * c * = o, (5) 

tn t + m 2 

and the loss of energy = ' l -\ *-S (6) 

2 2 

the entire energy being lost. 

§110. Special Cases of Elastic Impact. — (a) Let the 
mass m 2 be at rest. Then c 2 = o, 

2tn 2 c s ., 

v x = € x - 2 - L - Ci) 

m x -f- m 2 

2tn l c 1 , mi 

m l -f- tn z 



EXAMPLES OF ELASTIC AND INELASTIC IMPACT 1 29 

(b) Let m z be at rest, and let m 2 =00 . Then we have 
h = o, 

•'- V z = f, = <T X — 2^ = — ^, (3) 

z> 2 = o. (4) 

Hence the moving body retraces its path in the opposite direc- 
tion with the same velocity. 

(c) Let m 1 c l = —m 2 c 2 . Then our equations of condition 

become 

nijVj. -f- m 2 v 2 — o, 

«,p, 2 . m 2 v 2 m x c? . m~c 2 * 

1 = 1 : 

2222 

and from these we readily obtain 

t>i = —c x> 

v 2 = — c 2 \ 

i.e., both bodies return on their path with the same velocity 

with which they approached each other. 

§ in. Examples of Elastic and of Inelastic Impact. 

1. With what velocity must a body weighing 8 pounds strike one 
weighing 25 pounds in order to communicate to it a velocity of 2 feet 
per second, (a) when the bodies are perfectly elastic, (&) when wholly 
inelastic. 

2. Suppose sixteen impacts per minute take place between two bodies 
whose weights are respectively 1000 and 1200 pounds, their initial velo- 
cities being 5 and 2 feet per second respectively : find the loss of energy, 
the bodies being inelastic. 

§112. Imperfect Elasticity. — In this case we have the 
relations (see § 106) 

v — v t _ v 2 — v _ 
c x — v v — c 2 



I30 APPLIED MECHANICS. 



where 

and we have also 



__ m l c l + ni 2 c 2 . 
v — - , 

m t + m 2 



M 1 V l + ^2^2 = ^I^I + ^2^2* 

Determining from them the values of v s and v 2 , we obtain 
^ = v(i + <?) — **i, (i) 

^2 = »(i + e) — ec 2 , (2) 

or, by substituting for v its value, 

% = T (1 + *) - "i, (3) 

nil 4- ^2 

^ 2 = T — ( J + - "*• (4) 

*«i 4- m 2 

These may otherwise be put in the form 

v x = Cl -{i+e) m l (c t - c 2 ), (5) 
m x 4- m 2 

V 2 = c 2 + (1 4- e) ^ (V r - c 2 ). (6) 

/«i 4- m 2 

Moreover, we have for the loss of energy due to impact 

E = ^{c> - v*) + ^(c 2 2 - v 2 2 ) 
2 2 

or 

E = \\m x {c l - v l ){c 1 + v x ) 4- m 2 {c 2 — v 2 ){c 2 + v 2 )\; 

but, from (5) and (6) respectively, 

- 7 , _ (I + *)OT a fa - ^2) 

C x u x — 

w s -J- m 2 
, _ 7 , (1 4- g)«,(*« - fr) . 

f 2 — J/ 2 — 

#*i 4" W 2 



IMPERFECT ELASTICITY. 131 

m . m E = (1 +e){c, - c 2 ) ^ mm ^ €x + Vi) _ miM ^ 2 + ^1 

2(w t -f W 2 ) 

/. ^ = -r^-r^i - 0(1 + *)('i - '* + »i - *0- 

2(w, -f- m 2 ) 

But, from (i) and (2), 

0i — z> 2 = —e(c x — <: 2 ) 

2(/«! + W 2 ) 

or 

^ = (1 - «■) ;*'"'' , (*, - *)•■ (7> 

2{m 1 -f #z 2 ) 

When ^ = 1, or the elasticity is perfect, this loss of energy 
becomes zero. 

When e = o, or the bodies are totally inelastic, then the loss 
of energy becomes 

as has been already shown in § 107. 

An interesting fact in this connection is, that since (8) is 
the work expended in producing compression, and (7) is the 
work lost in all, therefore the work restored by the elasticity of 
the body is 



2 (m l -f- m 2 ) 



so that ^, or the square of the co-efficient of restitution, is the 
ratio of the work restored by the elasticity of the bodies, to 
the work expended in compressing the bodies up to the time 
of greatest compression. 



132 APPLIED MECHANICS. 

§113. Special Cases. — (a) Let m 2 be at rest, therefore 
c 2 = o. Then we shall have 

r.-VJi- *< I+ ;H » ,. »'-«»■, (1) 

( »*, + m 2 ) m t + m 2 

V, = (I + <)'. *'' , (2) 

and for loss of energy 

E= (1 -O ^ 2 *,«■ (3) 

(£) When m 2 = oo t and ^ 2 = o, we have 

z>x = -^1, (4) 

^2 = o, 

2? = (1 _ «.) ^i!. (5 ) 

2 



(c) When m 1 c l = —m 2 c 2y then 

z>i = — <*"„ 
^2 = -^2, 

jr, _ (1 — e 2 )m l c l (c 1 — c 2 ) _ (1 — e 2 )m z c 2 {c 2 — f r ) 
2 ' 2 

= (1 — <? 2 ) — - — — - — ^ ^ 2 . (6) 

2W, 



§114. Values of e as Determined by Experiment. — 
Since we have 

_ v — v t 

C — , 

C, — V 



IMPERFECT ELASTICITY. 133 

we shall have, when 

m 2 = oo and c 2 = o, 

m 1 c l + m 2 c 2 

v = — — = o. 

m x 4- m 2 

Hence 

Now, if we let a round ball fall vertically upon a horizontal 
slab from the height H y we shall have for the velocity of ap- 
proach 

c x = \[^B; 

and if we measure the height h to which it rises on its rebound, 
we shall have 

—v s = *J~2gh. 

Hence 



~5-vG 



H 

In this way the value of e can be determined experimentally 
for different substances. 

Newton found for values of e: for glass, ^| ; for steel, |; 
and Coriolis gives for ivory from 0.5 to 0.6. 

On the other hand, if we desired to adopt as our constant 
the ratio of the work restored, to the work spent in compres- 
sion, we should have for our constant *? 2 , and hence the squares 
of the preceding numbers. 



EXAMPLES. 

i. If two trains of cars, weighing 120000 and 160000 lbs., come 
into collision when they are moving in opposite directions with veloci- 
ties 20 and 15 feet per second respectively, what is the loss of mechan- 
ical effect expended in destroying the locomotives and cars ? 



134 APPLIED MECHANICS. 

2. Two perfectly inelastic balls approach each other with equal 
velocities, and are reduced to rest by the collision ; what must be the 
ratio of their weights ? 

3. Two steel balls, weighing .10 lbs. each, are moving with velocities 
5 and 10 feet per second respectively, and in the same direction : find 
their velocities after impact, the fastest ball being in the rear, and over- 
taking the other \ also the loss of mechanical effect due to the impact, 
assuming e — 0.55. 

§115. Oblique Impact. 

Let m„ m 2 , be the masses of the colliding bodies ; 
c„ c 2 , their respective velocities before impact ; 
a If a 2 , the angles made by c u c 2 , with the line of centres ; 
v x , v 2} the components of the velocities after impact ; 
c x cos a„ c 2 cos a 2 , the components of c lf c 2 , along the line of 

centres ; 
Cj sin a n c 2 sin a 2 , the components of c u c 2i at right angles to 

the line of centres ; 
v the common component of the velocity at the instant of 

greatest compression along line of centres ; 
z/, v", actual velocities after impact ; 
a, a", angles they make with line of centres ; 
v/ t v", actual velocities when compression is greatest ; 
a/, a/', angles they make with line of centres. 

Then we shall have, by proceeding in the same way as was done 
in §112, 

V x = c x cos a x — (1 + e) — (^cosa! — c 2 cos a^), (1) 

m x + m 2 



V 2 = C 2 COS a 2 -f- (i -|- e)- — * (l^COSaj — ^COScta), (2) 

m x -h m 2 



OBLIQUE IMPACT. 1 35 



if = \v t 2 4- <r I 2 sin 2 a I , 


(3) 


v" = >Jv 2 2 + c 2 2 sin 2 a,, 


(4) 


COS a = -„ 

V 


(S) 


COS a" = 3, 


(6) 


»?,<:, cos a, + m 2 c 2 C0Sa, 


(7) 


" — , » 


»/ = V^ 2 + <:■? sin 2 a„ 


(8) 



Ve " = Sjv 2 + ^ 2 sin 2 a,, (9) 

COS a/ = -^ ;, (io) 

V c 

COSa/' = ~. (il) 

And for the energy lost in impact, we have 

E = (I - ^)_-2i2?2L-— (^cosa, - ^COSa,)*. (l2> 

2(w I -f- m 2 ) 

When the bodies are perfectly elastic, 

e = 1, 
and equations (i), (2), and (12) become respectively 

2#Z 2 / \ 

V x = ^1 COS a x 2 (^ COS a x — <: 3 COS a,), 

# 2 = c 2 COS a 2 H ^^ — (^ COS a, — C 2 COS a,), 

.5 = 0. 

The rest remain the same in form. 

When the bodies are totally inelastic, 

e = o, 



136 APPLIED MECHANICS. 

and equations (1), (2), and (12) become respectively 

V x — C z COS a x (^COSa! — C 2 COS c^), 

v 2 = C 2 COS a 2 H (V x COS 04 — C 2 COS a,), 

E — — (^COSa! — ^COSOa) 2 . 

2(m l + m 2 ) 
The rest remain the same in form. 

§116. Impact of Revolving Bodies. — Let the bodies A 
and B revolve about parallel axes, and impinge upon each other. 
Draw a common normal at the point of contact. This 
common normal will be the line of impact. 
Let e x = angular velocity of A before impact, 
6 2 = angular velocity of B before impact, 
Wj = angular velocity of A after impact, 
<o 2 = angular velocity of B after impact, 
a x = perpendicular from axis of A on line of impact, 
a 2 = perpendicular from axis of B on line of impact, 
/, = moment of inertia of A about its axis, 
I 2 = moment of inertia of B about its axis. 

Then we shall have 

a I e I = Cj = linear velocity of A at point of contact before impact y 

a 2 z 2 = c 2 = linear velocity of B at point of contact before impact ; 

a l <D I = v t — linear velocity of A at point of contact after impact ; 

a 2 w 2 = v 2 = linear velocity of B at point of contact after impact ; 

I e 2 1 1 \c 2 

— — = ( — - )— = actual energy of A before impact ; 

*g W/ 2 g 

2 1 1 \c 2 
— = ( —2- ]— = actual energy of B before impact ; 

r \<* 2 2 /2g 

2 f I \v 2 

- = ( — - )— = actual energy of A after impact ; 
\a 2 )2g 

£-2- = ( — *- )— = actual energy of B after impact ; 
2g Wj2g &) v 



J, 

zg 

2? 






IMPACT OF REVOLVING BODIES. 1 37 

Hence it follows that we have the case explained in § 112 for 
imperfectly elastic impact, provided only we write 

— - instead of m z g and — instead of m 2 g. 
a* a 2 2 

Hence we shall have 

«, = C - «,(«,€, - a 2 c 2 ) Iz (1 + e), (1) 

/,#, -f- l % a x 

«, 2 = €2 + a 2 {a^ - a 2 € 2 )- -i— — (1 + e), (2) 

/> a * + Aa 2 2 

The case of perfect elasticity is obtained by making e = 1. 
The case of total lack of elasticity is obtained by making 
r — o. 

In the latter case the loss of energy is 

(a^ — g 2 c 2 ) 2 IJ 2 , v 

2^ />/ + 7 2 a/ V3; 

as can be seen by substituting the proper values in equation (8), 
§ 112. 



I38 APPLIED MECHANICS. 



CHAPTER III. 
ROOF-TRUSSES. 

§ 117. Definitions and Remarks The term "truss" may 

be applied to any framed structure intended to support a load. 

In the case of any truss, the external loads may be applied 
only at the joints, or some of the truss members may support 
loads at points other than the joints. 

In the latter case those members are subjected, not merely 
to direct tension or compression, but also to a bending-action, 
the determination of which we shall defer until we have studied 
the mode of ascertaining the stresses in a loaded beam ; and 
we shall at present confine ourselves to the consideration of 
the direct stresses of tension and compression. 

For this purpose any loads applied between two adjacent 
joints must be resolved into two parallel components acting at 
those joints, and the truss is then to be considered as loaded at 
the joints. By this means we shall obtain the entire stresses in 
the members whenever the loads are concentrated at the joints ; 
and, when certain members are loaded at other points, our re- 
sults will be the direct tensions and compressions of these mem- 
bers, leaving the stresses due to bending yet' to be determined. 

A tie is a member suited to bear only tension. 

A strut is a member suited to bear compression. 

§ 118. Frames of Two Bars. — Frames of two bars may 
consist, (1) of two ties (Fig. 69), (2) of two struts (Fig. 70), 
(3) of a strut and a tie (Fig. 71). 



FRAMES OF TWO BARS. 



1 39 



Case I. Two Ties (Fig. 69). — Let the load be repre- 
sented graphically by CF = W. 
Then if we resolve it into 
two components, CD and CE, 
acting along CB and CA re- 
spectively, CD will represent 
graphically the pull or tension 
in the tie CB, and CE that in 
the tie CA. 

The force acting on CB at 
B is equal and opposite to 
CD, while that acting on CA at A is equal and opposite to CE. 

To compute these stresses analytically, we have 




Fig. 69. 



CE = 



CD = 



Cj F sin CFE = W 



sin 1 



sin CEF sin(/ + *,) 



CF smCFD ^ x 



sin*! 



sin CDF sin(* -f t\) 



Case II. 7z£/^7 Struts (Fig. 




Fig. 70. 

C£ = W — sm ?I 



70). — Let the load be repre- 
sented graphically by CF= W. 
Then will the components CD 
and CE represent the thrusts 
in the struts CB and CA re- 
spectively, and the re-actions 
of the supports at B and A 
will be equal and opposite to 
them. For analytical solution, 
we derive from the figure 



sin(/ + t\Y 



CD = W 



smz 



sin(z" -j- i x ) 



Case III. A Strut and a Tie (Fig. 71). — Let the load be 
represented graphically by CF = W. Resolve it, as before, 
into components along the members of the truss. Then will 



140 



APPLIED MECHANICS. 




Fig. 71. 



CE represent the tension in the tie AC, and CD will represent 
the thrust in the strut BC ; and we may 
deduce the analytical formulae as before. 

§ 119. Stability for Lateral Deviations. 
— In Case I, if the joint C be moved a little 
out of the plane of the paper, the load at 
C has such a direction that it will cause the 
truss to rotate around AB so as to return to 
its former position ; hence such a frame is 
stable as regards lateral deviations. 

In Case II the effect of the load, if C 
were moved a little out of the plane of the 
paper, would be to cause rotation in such a way as to overturn 
the truss ; hence such a frame is unstable as regards lateral 
deviations. 

In Case III the stability for lateral deviations will depend 
upon whether the load CF == W is parallel to AB, is directed 
away from it or towards it. If the first is the case (i.e., if A is 
the point of suspension of the tie), the frame is neutral, as the 
load has no effect, either to restore the truss to its former posi- 
tion, or to overturn it ; if the second is the case (i.e., if A t is 
the point of suspension of the tie), the truss is stable ; and, if 
the third is the case (i.e., if A 2 is the point of suspension of the 
tie), it is unstable as regards lateral deviations. 

§ 120. General Methods for Determining the Stresses in 
Trusses. — In the determination of the stresses as above, it 
would have been sufficient to construct only the triangle CFD 
by laying off CF = IV to scale, and then drawing CD parallel 
to CZ?/and FD parallel to CA, and the triangle CFD would have 
given us the complete solution of the problem. Moreover, the 
determination of the supporting forces of any truss, and of the 
stresses in the several members, is a question of equilibrium. 
Adopting the following as definitions, viz., — 
External forces are the loads and supporting forces, 



TRIANGULAR FRAME. I4I 

Internal forces are the stresses in the members : 
we must have 

i°. The external forces must form a balanced system; i.e., 
the supporting forces must balance the loads. 

2 . The forces (external and internal) acting at each joint 
of the truss must form a balanced system ; i.e., the external 
forces (if any) at the joint must be balanced by the stresses in 
the members which meet at that joint. 

3 . If any section be made, dividing the truss into two parts, 
the external forces which act upon that part which lies on one 
side of the section, must be balanced by the forces (internal) 
exerted by that part of the truss which lies on the other side 
of the section, upon the first part. 

The above three principles, the triangle, and polygon of 
forces, and the conditions of equilibrium for forces in a plane, 
enable us to determine the stresses in the different members 
of roof and bridge trusses. 

§121. Triangular Frame. — Given the triangular frame 
ABC (Fig. 72), and given the load W at C in magnitude and 
direction, given also the 
direction of the support- 
ing force at B, to find the 
magnitude of this support- 
ing force, the magnitude 
and direction of the other 
supporting force, and the 
stresses in the member?. 

Solution. — Join A 
with D, the point of inter- Fig. 72. 

section of- the line of direction of the load and the line BE. 
Then will DA be the direction of the other supporting force ; 
for the three external forces, in order to form a balanced sys- 
tem, must meet in a point, except when they are parallel. 
Then draw ab to scale, parallel to CD and equal to W. From 




142 



APPLIED MECHANICS. 



a draw ac parallel to BD, and from b draw be parallel to AD ; 
then will the triangle abca be the triangle of external forces, 
the sides ab, be, and ca, taken in order, representing respectively 
the load W, the supporting force at A, and the supporting force 
at B. 

Then from a draw ad parallel to BC, and from c. draw cd 
parallel to AB ; then will the triangle acd be the triangle of 
forces for the joint B, and the sides ca, ad, and dc, taken in 
order, will represent respectively the supporting force at B, the 
force exerted by the bar BC at the point B, and the force 
exerted by the bar AB at the point B. 

Since, therefore, the force ad exerted by the bar CB at B 
is directed away from the bar, it follows that CB is in compres- 
sion ; and, since the force dc exerted by the bar AB at B is 
directed towards the bar, it follows that AB is in tension. 

In the same way bde is the triangle of forces for the point 
A ; the sides be, cd, and db representing respectively the sup- 
porting force at A, the force 
exerted by the bar AB at A, 
and the force exerted by the 
bar AC at A. 

The bar AB is again seen to 
be in tension, as the force cd 
exerted by the bar AB at A is 
directed towards the bar. 

So likewise the triangle abd 
is the triangle of forces for the 
point C 

Fig. 73 shows the case when 
the supporting forces meet the load-line above, instead of 
below, the truss. 

§ 122. Triangular Frame with Load and Supporting 
Forces Vertical. — Fig. 74 shows the construction when the 
load and also the supporting forces are vertical. In this case 




BOW'S NOTATION. 



143 




Of 



Fig. 74. 

the stress diagrams of roof- 



the diagram becomes very much simplified, the triangle of 
external forces abd becom- 
ing a straight line. The 
diagram is otherwise con- 
structed just like the last 
one. 

§ 123. Bow's Notation. 
— The notation devised by 
Robert H. Bow very much 
simplifies the construction 
trusses. 

This notation is as follows : Let the radiating lines (Fig. 75) 
represent the lines of action of a system of forces in equilib- 
rium, and let the polygon abcdefa be the polygon representing 

these forces in magnitude 
and direction ; then denote 
the sides of the polygon 
in the ordinary way, by 
placing small letters at the 
vertices, but denote the 
radiating lines by capital 
letters placed in the angles. 
Thus the line AB is the 
line of direction of the 
force ab, etc. In applying the notation to roof-trusses, we letter 
the truss with capital letters in the spaces, and the stress dia- 
gram with small letters at the vertices. If, then, in drawing 
the polygon of equilibrium for any one joint of the truss, we 
take the forces always in the same order, proceeding always 
in right-handed or always in left-handed rotation, we shall be 
led to the simplest diagrams. Hereafter this notation will be 
used exclusively in determining the stresses in roof-trusses. 

§124. Isosceles Triangular Frame: Concentrated Load 
(Fig. 76.) — Let the load W act at the apex, the supporting 




Fig. 75. 



144 



APPLIED MECHANICS. 




Fig. 76. 



forces being vertical ; each will be equal to \W : hence the 
polygon of external forces will be the triangle abc, the sides of 

which, ab y be, and ca, all lie in 
one straight line. Then begin 
at the left-hand support, and 
proceed again in right-handed 
rotation, and we have as the tri- 
angle of forces at this joint cad, 
the forces ca, ad, and dc t these 
being respectively the support- 
ing force, the stress in AD, and 
that in DC; the directions of 
these forces being indicated by 
the order in which the letters follow each other : thus, ca is an 
upward force, ad is a downward force ; and, this being the 
force exerted by the bar AD at the left-hand support, we con- 
clude that the bar AD is in compression. Again : dc is 
directed towards the right, or towards the bar itself, and hence 
the bar DC is in tension. The triangle of forces for the other 
support is bed, and that for the apex abd. 

§ 125. Isosceles Triangular Frame: Distributed Load. — 
Let the load W be uniformly distributed over the two rafters 
AF and FB (Fig. 77) ; then will 
these two rafters be subjected to 
a direct stress, and also to a bend- 
ing action : and if we resolve the 
load on each rafter into two com- 
ponents at the ends of the rafter, 
then, considering these components 
as the loads at the joints, we shall 
determine correctly by our diagram the direct stresses in all 
the bars of the truss. 

The load distributed over AF is — ; and of this, one-half is 




Fig. 77. 



POLYGONAL FRAME. 



145 



the component at the support, and one-half at the apex, and 

similarly for the other rafter. This gives as our loads, — at 

4 

each support, and — at the apex. The polygon of external 
2 

forces is eabcde, where the sides are as follows : — 



W 



W 

2 



be = 



W 



cd = — , 



de = 



W 



Then, beginning at the left-hand support, we shall have for the 
polygon of forces the quadrilateral deafd, where de = — = sup- 
porting force, ea = — = downward load at support, af = 

4 
stress in AF (compression), fd = stress in FD (tension). The 
polygon for the apex is abf y and that for the right-hand support 
cdfbc. 

§ 126. Polygonal Frame. — Given a polygonal frame (Fig. 
78) formed of bars jointed together at the vertices of the angles, 
and free to turn on these joints, 
it is evident, that, in order that 
the frame may retain its form, 
it is necessary that the direc- 
tions of, and the proportions 
between, the loads at the dif- 
ferent joints, should bt speci- 
ally adapted to the given form : 
otherwise the frame will change 
its form. We will proceed to 
solve the following problem : 
Given the form of the frame, 
the magnitude of one load as AB, and the direction of all the 
external forces (loads and supporting forces) except one, we 
shall have sufficient data to determine the magnitudes of all, 




Fig. 78. 



146 



APPLIED MECHANICS. 



and the direction of the remaining external forces, and also the 
stresses in the bars 

Let the direction of all the loads be given, and also that of 
the supporting force EF, that of the supporting force AF being 
thus far unknown ; and let the magnitude of AB be given. 
Then, beginning at the joint ABG, we have for triangle of 
forces abg formed by drawing ab || and = AB, then drawing 
ga || AG, and bg || BG ; ga and bg both being thrusts. Then, 
passing to the joint BCG, we have the thrust in BG already- 
determined, and it will in this case be represented by gb. If, 
now, we draw be || BC, and gc || GC, we shall have determined 
the load BC as be, and we shall have eg and gb as the thrusts 
in CG and GB respectively. Continuing in the same way, we 
obtain the triangles gcd, gde, and gfe, thus determining the 
magnitudes of the loads cd, de, and of the supporting force ef; 
and then the triangle ga/,. formed by joining a and/, gives us of 
for the magnitude and direction of the left-hand support. The 
polygon abedefa of external forces is called the Force Polygon, 
while the frame itself is called the Equilibrium Polygon. 

§ 127. Polygonal Frame with Loads and Supporting 
Forces Vertical. — In this case (Fig. 79) we may give the 

form of the frame and the mag- 
nitude of one of the loads, to 
determine the other loads and 
the supporting forces, and also 
the stresses in the bars ; or we 
may give the form of the frame 
and the magnitude of the re- 
sultant of the loads, to find the 
loads and supporting forces. In 
the former case let the load AB 
be given. Then, proceeding in 
the same way as before, we find the diagram of Fig. 79 ; the 
polygon of external forces abedefa falling all in one straight line. 




Fig. 79. 



FUNICULAR POLYGON. — TRIANGULAR TRUSS. 



147 



If, on the other hand, the whole load ae be given, we observe 
that this is borne by the stresses in the extreme bars AG and 
GE ; hence, drawing ag || AG, and eg || EG, we find eg and ga 
as the. stresses in EG and GA respectively. Then, proceeding 
to the joint ABG, we find, since ga is the force exerted by 
GA at this point, that, drawing gb || GB, we shall have ab as 
the part of the load acting at the joint ABG y etc. 

§ 128. Funicular Polygon. — If the frame of Fig. 79 be 
inverted, we shall have the 
case of Fig. 80, where all 
the bars, except EG, are sub- 
jected to tension; EG itself 
being subjected to compres- 
sion. The construction of the 
diagram of stresses being en- 
tirely similar to that already 
explained for Fig. 79, the ex- 
planation will not be repeated 
here. If the compression 
piece be omitted, the case 
becomes that of a chain hung 
at the upper joints (the supporting forces then becoming iden- 
tical with the tensions in the two extreme bars), the line gf 
would then be omitted from the diagram, and the polygon of 
external forces would become abcdega. 

§ 129. Triangular Truss : Wind Pressure. — Inasmuch as 
the pressure of the wind on a roof has been shown by experi- 
ment to be normal to the roof on the side from which it blows, 
we will next consider the case of a triangular truss with the 
load distributed over one rafter only, and normal to the rafter. 




Fig. 80. 



There may be three cases : — 

i°. When there is a roller under one end, and the wind 
blows from the other side. 



148 



APPLIED MECHANICS. 



2°. When there is a roller under one end, and the wind 
blows from the side of the roller. 

3 . When there is no roller under either end. 

The last arrangement should always be avoided except in 
small and unimportant constructions ; for the presence of a 
roller under one end is necessary to allow the truss to change 
its length with the changes of temperature, and to prevent the 
stresses that would occur if it were confined. 



Case I. — Using Bow's notation, we have (Fig. 81) the 

whole load represented 
in the diagram by db. 
Its resultant acts at the 
middle of the rafter 
AE, whereas the sup- 
porting force at the 
right-hand end is (in 
consequence of the pres- 
ence of the roller) verti- 
cal. Hence, to find the 
line of action of the other 
supporting force, pro- 
duce the line of action 
of the load till it meets 
a vertical line drawn 

through the roller, and join their point of intersection with the 

support where there is no roller. We thus obtain CD as the 

line of action of the left-hand support. 

We can now determine the magnitude of the supporting 

forces be and cd by constructing the triangle bedb of external 

forces. 

Now resolve the normal distributed force db into two single 

forces (equal to each other in this case), da and ab respectively, 

acting at the left-hand support and at the apex. 




Fig. 81. 



TRIANGULAR TRUSS: WIND PRESSURE. 



I49 



Now proceed to the left-hand support. We find four forces 
in equilibrium, of which two are entirely known ; viz., cd and 
da: hence, constructing the quadrilateral cdaec, we have ae as 
the thrust in AE, and ec as the tension in EC. 

Next proceed to the apex, and construct the triangle of 
equilibrium abea, and we obtain be as the thrust in BE. 

The triangle bceb is then the triangle of equilibrium for the 
right-hand sup- 
port. 

Case II. — 
In this case 
(Fig. 82) we fol- 
low the same 
method of pro- 
cedure, only the 
point of inter- 
section of the 
load and sup- 
porting forces 

is above, instead of below, the truss. The figure explains itself 

so fully that it is unnecessary to 
explain it here. 



Case III. — In this case the 
supports are capable of exerting 
resistance in any direction what- 
ever ; so that, if any circumstance 
should determine the direction 
of one of them, that of the other 
When there is no such circum- 




Fig. 82. 




Fig. 83. 



would be determined also 
stance, it is customary to assume them parallel to the load 
(Fig. 83). Making this assumption, we begin by dividing the 
line db, which represents the load, into two parts, inversely 



150 APPLIED MECHANICS. 

proportional to the two segments into which the line of action 
of the resultant of the load (the dotted line in the figure) 
divides the line EC. We thus obtain the supporting forces be 
and cd, and bedb is the triangle of external forces. We then 
follow the same method as in the preceding cases. 

§ 130. General Determination of the Stresses in Roof- 
Trusses. — In order to compute the stresses in the different 
members of a roof-truss, it is necessary first to know the 
amount and distribution of the load. 

This consists generally of — 

i°. The weight of the truss itself. 

2 . The weight of the purlins, jack-rafters, and superin- 
cumbent roofing, as the planks, slate, shingles, felt, etc. 

3 . The weight of the snow. 

4 . The weight of the ceiling of the room immediately 
below if this is hung from the truss, or the weight of the 
floor of the loft, and its load, if it be used as a room. 

5°. The pressure of the wind ; and this may blow from 
either side. 

6°. Any accidental load depending on the purposes for which 
the building is used. As an instance, we might have the case 
where a system of pulleys, by means of which heavy weights 
are lifted, is attached to the roof. 

In regard to the first two items, and the fourth, whenever 
the construction is of importance, the actual weights should 
be determined and used. In so doing, we can first make an 
approximate computation of the weight of the truss, and use it 
in the computation of the stresses ; the weights of the ceiling 
or of the floor below being accurately determined. After the 
stresses in the different members have been ascertained by the 
use of these loads, and the necessary dimensions of the mem- 
bers determined, we should compute the actual weight of the 
truss ; and if our approximate value is sufficiently different 
from the true value to warrant it, we should compute again 



STRESSES IN ROOF-TRUSSES. 



151 



the stresses. This second computation will, however, seldom be 
necessary. 

In making these computations, the weights of a cubic foot 
of the materials used will be needed ; and average values are 
given in the following table with sufficient accuracy for the 
purpose. 



Weight of some Building Ma- 
terials per Cubic Foot. 



Pounds. 



Weight of Slating per Square 

Foot. 

According to Trautwine. 



Pounds. 



Timber. 

Chestnut 

Hemlock 

Maple 

Oak, live 

Oak, white 

Pine, white 

Pine, yellow, Southern 
Spruce 

Iron. 

Iron, cast 

Iron, wrought 

Steel 

Other Substances. 

Asphaltum 

Mortar, hardened . . . 
Snow, freshly fallen . . 
Snow, compacted by rain . 
Slate 



41 
25 
41 
59 
49 
25 to 30 

45 
25 to 30 



45° 
480 
490 



80 to 90 
103 
5 to 12 

15 to 5° 
140 to 180 



\ inch thick on laths 



i " 

A" 

tV' 



i-inch boards 
ii " 

laths . . 
i-inch boards 
\\ " 

laths . . 
i-inch boards, 
i£ " 



With slating-felt add . 
With £-inch mortar add 



Number of Nails in One Pound. 



3-penny 

4 

6 
8 

TO 
12 

20 



475 
6.75 
7.30 
7.00 
9.00 

9-55 
9.25 
11.25 
11.80 
±lb. 
3 lbs. 



No. 



450 
340 

150 

100 

60 

40 
25 



As to the weight of the snow upon the roof, Stoney recom- 
mends the use of 20 pounds per square foot in moderate 
climates ; and this would seem to the writer to be borne out by 
the experiments of Trautwine as recorded in his handbook, 



I5 2 APPLIED MECHANICS. 

although Trautwine himself considers 12 pounds per square 
foot as sufficient. 

§ 131. Wind Pressure. — While a great deal of work has 
been done to ascertain the direction and the greatest intensity 
of the pressure of the wind upon exposed surfaces, as those of 
roofs and bridges, nevertheless the amount of information on 
the subject is very small, inasmuch as but few experiments 
have been under the conditions of practice. Before giving a 
summary of what has been done the following statements will 
be made : 

i°. The pressure of the wind upon a roof, or other surface, 
is assumed to be normal to the surface upon which it blows ; 
and what little experimenting has been done upon the subject 
tends to confirm this view. 

2°. Inasmuch as more attempts have been made to deter- 
mine experimentally the velocity of the wind than its pressure, 
hence there have been a good many experiments to determine 
the relation between the velocity and the pressure upon a sur- 
face to which the direction of the wind is normal. 

3 . A few experimenters have tried to determine the rela- 
tion between the intensity of the pressure on a surface normal 
to the direction of the wind and one inclined to its direction. 

4 . While the above have been the investigations most com- 
monly pursued, other subjects of experiment have been — 

(a) The variation of pressure with density ; (b) with tem- 
perature ; (c) with humidity ; (d) with the size of surface 
pressed upon ; (e) with the shape of surface pressed upon ; 
(/) whether the pressure corresponding to a certain velocity is 
the same whether the air moves against a body at rest, or 
whether the body moves in quiet air. 

By way of references to the literature of the subject may 
be given the following, as most of the work that has been 
done is included in them or in other references which they 
contain : 



WIND PRESSURE. 1 53 



i°. Proceedings of the British Institution of Civil Engineers, vol. 
lxix., year 1882, pages 80 to 218 inclusive. 

2 . A. R. Wolff : Treatise on Windmills. 

3 . C. Shaler Smith : Proceedings American Society of Civil En- 
gineers, vol. x., page 139. 

4 . A. L. Rotch : Report of Work of the Blue Hill Meteorological 
Observatory, 1887. 

5 . Engineering, Feb. 28th, 1890 : Experiments of Baker. 

6°. Engineering, May 30, June 6, June 13, 1890 : Experiments of 
O. T. Crosby. 

The first gives an account of a very full discussion of the 
subject, by a large number of Engineers. The second con- 
tains a recommendation that the temperature of the air be con- 
sidered in estimating the pressure. The fifth gives an account 
of Baker's experiments on wind pressure in connection with the 
building of the Forth Bridge. 

Before an account is given of the experimental work that 
has been done, the following statements will be made of what 
are some of the methods in most common use : 

1°. A great many engineers very commonly call from 40 to 
55 pounds per square foot the maximum pressure on a vertical 
surface at right angles to the direction of the wind. One rather 
common practice, in the case of bridges, is to estimate 30 
pounds per square foot on the loaded, or 50 pounds per square 
foot on the unloaded structure. Nevertheless pressures of 80 
and 90 pounds per square foot have been registered and re- 
corded by the use of small pressure-plates, and by computation 
from anemometer records. 

2°. By way of determining the intensity of the pressure on 
an inclined surface in terms of that on a surface normal to the 
direction of the wind, four methods more or less used will be 
enumerated here : 

(a) Duchemin's formula, which Professor W. C. Unwin 
recommends, is as follows, viz. : 



154 



APPLIED MECHANICS. 



P=A 



2 sin 



6 



i + sin 2 0' 



where p = intensity of normal pressure on roof, p x = intensity 
of piessure on a plane normal to the direction of the wind. 
(b) Hutton's formula, 

p=p x (sin 0)i.84 cos 0-^ 

Unwin claims that this and Duchemin's formula give nearly 
the same results for all angles of inclination greater than 15 . 

The following table gives the results obtained by the use of 
each, on the assumption that p Y = 40 : 



6 


Duchemin. 


Hutton. 





Duchemin. 


Hutton. 


5° 


6.89 


5-IO 


50° 


38.64 


38.10 


IO° 


13-59 


9.60 


55° 


39.21 


39-40 


15° 


19.32 


14.20 


60 ° 


39-74 


40.00 


20° 


24.24 


18.40 


65° 


39-82 


40.00 


25° 


28.77 


22.6o 


7o° 


39-91 


40.00 


30° 


32.00 


26.50 


75° 


39 -9 6 


40.00 


35° 


34-52 


30.IO 


8o° 


40.00 


40.00 


40° 


36.40 


33-30 


85° 


40.00 


40.00 


45° 


37-73 


36.OO 


90 


40.00 


40.00 



(c) A formula very commonly favored, but which does not 
agree with any experiments that have been made, is 

p = p x sin 2 d. 

It gives much lower results, as a rule, than either of the others, 
but it is favored by many because, if we assume the wind to 
blow in parallel lines till it strikes the surface, and then to get 
suddenly out of the way, forming no eddies on the back side 
of the surface and meeting no lateral resistance on the front 



WIND PRESSURE. I 55 



side, all of which are conditions that do not exist, we could 
then deduce it as follows: 

Assume a unit surface making an angle 6 with the direction 
of the wind, the total pressure on this surface in the direction 
of the wind would be p x sin 0; and by resolving this into nor- 
mal and tangential components we should have, for the former, 

p = p 1 sin 3 6. 

{d) Another rule which is sometimes used, but which has 
nothing to recommend it, is to consider the normal intensity 
of the wind pressure per square foot of roof surface as equal to 
the number of degrees of inclination of the roof to the hori- 
zontal. The wind pressure allowed for by this rule is very 
excessive, as it would be 90 pounds per square foot for a ver- 
tical surface. 

Taking up, now, the experimental work that has been done, 
we will begin with the attempts to determine velocities and 
pressures, and the relation between them. 

i°. In regard to velocities, these are determined by using 
some kind of an anemometer, and in all these cases there are 
several difficulties and sources of error, as follows : 

(a) In many cases the anemometers have not even been 
graduated experimentally, but it has been assumed outright 
that the velocity of the air is just three times the linear velocity 
of the cups of a cup anemometer. 

ip) When they have been graduated, it has generally been 
done by attaching them to the end of the arm of a whirling 
machine, which, when the arm is long, and the velocity moder- 
ate, will do very well, but is the more inaccurate the shorter 
the arm and the higher the velocity of motion. 

(c) The wind always comes in gusts, and hence the ane- 
mometer does not register the average velocity of the wind at 
any one moment, but that of the particular portion that comes 



156 APPLIED MECHANICS. 

in contact with it, and this is always a small portion, on ac- 
count of the small size of the anemometer. 

(d) In order to get an indication which is not affected by 
the cross-currents reflected from the surrounding buildings and 
chimneys, it is necessary to put the anemometer very high up, 
and then, of course, we obtain the indications corresponding 
to that height, which is greater than that of the buildings, and 
it is well known that the velocity of the wind increases very 
considerably with the height. 

Next, as to the direct determination of pressure, this has 
usually been done by means of some kind of pressure-plate, 
either round or square, but of small size, thus allowing the 
eddies formed on the back side of the plate to have a con- 
siderable effect. The results obtained by the use of different 
sizes and different shapes of plates have therefore differed very 
considerably ; and while some have claimed that the pressure 
per square foot increases with the size of the surface pressed 
upon, it has been very thoroughly proved by the more modern 
investigations that the opposite is true, and that the pressure 
decreases with the size. 

While the records from small pressure-plates have fre- 
quently shown very high pressures per square foot, as 80, 90, 
or even over 100 pounds per square foot, it has become very 
generally recognized by engineers that by far the greater part 
of existing buildings and bridges would be overturned by winds 
of such force, or anywhere near such force, and it has not been 
customary among them to make use of such high figures for 
wind pressure on bridges and roofs in computing the stability 
of structures. While some of the figures in general use have 
already been given, nevertheless the tendency of modern inves- 
tigation seems to be to obtain rather lower figures. In this con- 
nection it is well to refer to the work done by Baker in connec- 
tion with the construction of the Forth Bridge. The following 
description is taken from " Engineering" of Feb. 28th, 1890: 



WIND PRESSURE. 



157 



" The wind pressure to be provided for in the calcu- 
lations for bridges in exposed positions is 56 lbs. per square 
foot, according to the Board of Trade regulations, and this 
twice over the whole area of the girder surface exposed, the 
resistance to such pressure to be by dead-weight in the struc- 
ture alone. 

" The most violent gales which have occurred during the 
construction of the Forth Bridge are given, with the pressures 
recorded on the wind gauges, in the annexed table : 





Month 
and 
Day. 


Pressure in pounds per square foot. 




Year. 


Revolving 
Gauge. 


Small 

fixed 

Gauge. 


Large 

fixed 

Gauge. 


In centre 
of large 
Gauge. 


Right- 
hand top 
of large 
Gauge. 


Direction 

of 

Wind. 


1883 
1884 
1884 
1884 
1885 
1885 
1886 
1887 
1888 
1888 
1889 
1890 
1890 
1890 


Dec. II, 

Jan. 26, 

Oct. 27, 

Oct. 28, 

Mar. 20, 

Dec. 4, 

Mar. 31, 

Feb. 4, 

Jan. 5, 

Nov. 17, 

2, 

Jan. 19, 

" 21, 

" 22, 


33 
65 
29 
26 
30 
25 
26 
•26 
27 
35 
27 
27 
26 
27 


39 
41 
23 
29 

25 
27 

31 
41 
16 
41 

34 
28 
38 
24 


22 

35 
18 

19 
17 
19 
19 
15 
7 
27 
12 
16 

15 

18 


23i 


22 


s. w.* 
s. w.* 
s. w. 
s. w. 

w. 

w. 
s. w. 
s. w. 

S. E. 

w. 

s. w. 
s. w. 

w. 

S. W. by W. 



* These data are unreliable, owing to faulty registration by the indicator-needle, as will 
b* presently explained. They were altered after this date. The barometer fell to 27.5 inches 
on <hat occasion, over three quarters of an inch within an hour. 



158 APPLIED MECHANICS. 

"The pressure-gauges, which were put up in the summer 
of 1882 on the top of the old castle of Inchgarvie, and from 
which daily records have been taken throughout, were of very- 
simple construction. The maximum pressures only were taken. 
The most unfavorable direction from which the wind pressure 
can strike the bridge is nearly due east and west, and two out 
of the three gauges were fixed to face these directions, while 
a third was so arranged as to register for any direction of 
wind. 

"The principal gauge is a large board 20 feet long by 15 
feet high, or 300 square feet area, set vertically with its faces east 
and west. The weight of this board is carried by two rods sus- 
pended from a framework surrounding the board, and so ar- 
ranged as to offer as little resistance as possible to the passage 
of the wind, in order not to create eddies near the edge of the 
board. In the horizontal central axis of the board there are 
fixed two pins which fit into the lower eyes of the suspension- 
rods, the object being to balance the board as nearly as pos- 
sible. Each of the four corners of the board is held between 
two spiral springs, all carefully and easily adjusted so that any 
pressure exerted on either face will push it evenly in the op- 
posite direction, but upon such pressure being removed the 
compressed springs will force the board back to its normal 
position. To the four corners four irons are attached, uniting 
in a pyramidal formation in one point, whence a single wire 
passes over a pulley to the registering apparatus below. In 
order to ascertain to some extent how far great gusts of wind 
are quite local in their action, and exert great pressure only 
upon a very limited area, two circular spaces,. one in the exact 
centre and one in the right-hand top corner, about 18 inches 
in diameter, were cut out of the board and circular plates in- 
serted, which could register independently the force of the 
wind upon them. 

" By the side of the large square board, at a distance of 



WIND PRESSURE. I 59 



about 8 feet, another gauge, a circular plate of 1.5 square feet 
area, facing east and west, was fixed up with separate regis- 
tration. This was intended as a check upon the indications 
given by the large board. 

" Another gauge of the same dimensions as the last, but 
with the disc attached to the short arm of a double vane, so 
that it would face the wind from whatever direction it might 
come, was set up. 

" On one occasion the small fixed board appeared to regis- 
ter 65 pounds to the square foot — a registration which caused 
no little alarm and anxiety. Mr. Baker found, upon inves- 
tigation, that the registering apparatus was not in good order, 
and after adjusting it the highest pressure recorded was 41 
pounds. 

" In order to determine the effect of the wind upon surfaces 
like that of the exposed surface of the bridge, he devised an 
apparatus which consisted of a light wooden rod suspended in 
the middle, so as to balance correctly, by a string from the 
ceiling. At one end was attached a cardboard model of the 
surface, the resistance of which was to be tested, and at the 
opposite end was placed a sheet of cardboard facing the same 
way as the model, so arranged that by means of another and 
adjustable sheet, which would slide in and out of the first, 
the surface at that end could be increased or decreased at 
the will of the operator. The mode of working is for a 
person to pull it from its perpendicular position towards 
himself, and then gently release it, being careful to allow 
both ends to go together. If this is properly done, it is evi- 
dent that the rod will in swinging retain a position parallel 
to its original position, supposing that the model at one 
end and the cardboard frame at the other are balanced as 
to weight, and that the two surfaces exposed to the air 
pressure coming against it in swinging are exactly alike. 
Should one area be greater than the other, the model or card- 



l60 APPLIED MECHANICS. 



board sheet, whichever it may be, will b~ lagging behind, and 
twist t>e string." 

The experiments carried on in various ways by different 
people and at different times are generally in agreement with 
each other and with the results of more elaborate processes. 
The information specially desired was in regard to the wind 
pressure upon surfaces more or less sheltered by those imme- 
diately in front of them. In this regard Mr. Baker satisfied 
himself that, while the results differed very considerably ac- 
cording to the distance apart of the surfaces, in no case was 
the area affected by the wind, in any girder which had two or 
more surfaces exposed, more than 1.8 times the area of the 
surface directly fronting the wind, and, as the calculations had 
been made for twice this surface, the stresses which the struc- 
ture will receive from this cause will be less than those pro- 
vided for. 

Next, as to the relation between velocity and pressure, a 
great many formulae have been devised, to agree with the 
results of different experimenters. Most all of them make the 
pressure proportional to the square of the velocity ; while 
some add a term proportional to the velocity itself, and when 
higher velocities are reached, as those usual in gunnery, terms 
have been introduced with powers of the velocity higher than 
the second. It is hardly worth while to consider these dif- 
ferent formulae, as it is rather the pressure than the velocity 
that the engineer is interested in, and correct information in 
this regard is to be obtained rather from pressure-boards than 
from anemometers. Nevertheless, it may be stated that one 
of the most usual formulae is that of Smeaton,- and is 

200 
where P— pressure in pounds per square foot, and V= velocity 



WIND PRESSURE. l6l 



in miles per hour. This formula agrees very well with a num- 
ber of experiments that have been made where anemometers 
have been used to determine the velocity, and small pressure- 
plates (say one square foot) to determine the pressure ; thus 
this formula satisfies very well the experiments made at the 
Blue Hill Meteorological Observatory, near Boston, Mass., 
U. S. A. 

It was originally deduced from some very old experiments 
of Rouse ; and it agrees with a good many, but disagrees with 
other experiments. It is probably the formula that has been 
more quoted than any other. 

A little ought also to be said in regard to the pressure of 
the wind on very high structures, as on the piers of high via- 
ducts and on tall chimneys. In this regard it is to be ob- 
served : 

i°. The pressure, as well as the velocity of the wind, be- 
comes greater the higher up from the ground the surface ex- 
posed is situated. 

2°. From calculations on chimneys that have stood for a 
long time, Rankine deduced, as the greatest average wind 
pressure that could be realized in the case of tall chimneys, 55 
pounds per square foot. 

3 . In making the piers of high viaducts, it would seem 
desirable not to make them solid, but to use only four up- 
rights at the corners connected by lattice work, in order to 
expose a smaller surface to the wind. Nevertheless, as was ex- 
plained, it will not do to separate the structure into its com- 
ponent parts, and to estimate the pressure on each part 
separately and then add the results together to get the total 
effect ; but we really need some such experiments as those of 
Baker. 

4 . Some old experiments of Borda bear out the common 
practice of assuming the wind pressure on the surface of a cir- 



1 62 APPLIED MECHANICS 

cular cylinder one half that which would exist on its projection 
on a plane normal to the direction of the wind. 

There remains now only to refer to a serial article by O. T. 
Crosby, in " Engineering" of May 30, June 6th, and June 13th, 
1890, containining some experiments made by him on wind 
pressure near Baltimore, Md. The first two numbers contain 
rather a summary of what has been done by others, and it is 
in the copy of June 13th that is to be found the account of his 
own work, which was done in order to determine the resistances 
of the air to fast-moving trains. 

He used a whirling arrangement turning about a vertical 
axis, to the end of which was attached a car, the circumference 
through which the car moved being 36 feet. 

In order to determine whether the circular motion produced 
any disturbing effect, he ran a car having a cross-section of 5.1 
square feet on a circular track about two miles in circumference, 
the speed of the car being about 50 miles per hour, and the 
results obtained in this way agreed very nearly with those ob- 
tained from his whirling table. The special peculiarity of his 
results is that he obtained, by plotting them, the law that the 
pressure varies directly as the first power of the velocity, and 
not as the square or some higher power; also, his pressures, 
after the velocity had passed 25 or 30 miles per hour, are 
much lower than those given by Smeaton and others, the pres- 
sure on a normal plane surface moving at 115 miles per hour 
being about 27 pounds per square foot. 

The cars used were generally about 3 feet long without the 
front. The fronts attached were: i°. Normal plane surface; 
2°. Wedge, base 1, height 1 ; 3 . Pyramid, base 1, height 2; 4 . 
Wedge and cyma, base 1, height 2; 5 . Parabolic wedge, 
base 1, height 2. 

His experiments covered a range of velocities from 30 to 
130 miles per hour. 



DISTRIBUTION OF THE LOADS. 1 63 

The law of the first powers of the velocities seems peculiar, 
and certainly ought not to be accepted without further cor- 
roborative evidence ; but the low values of the pressures agree 
with Baker's results and with the tendency of the more modern 
investigations. 

§ 132. Approximate Estimation of the Load. — In all 
important constructions, the estimates of the loads should be 
made as above described. For smaller constructions, and for 
the purposes of a preliminary computation in all cases, we only 
estimate the roof-weight roughly ; and some authors even as- 
sume the wind pressure as a vertical force. 

Trautwine recommends the use of the following figures for 
the total load per square foot, including wind and snow, when 
the span is 75 feet or less : — 

Roof covered with corrugated iron, unboarded - . . . 28 lbs. 

Roof plastered below the rafters . . 38 " 

Roof, corrugated iron on boards 31 " 

Roof plastered below the rafters 41 " 

Roof, slate, unboarded or on laths . '. 33 " 

Roof, slate, on boards \\ inches thick 35 " 

Roof, slate, if plastered below the rafters 46 " 

Roof, shingles on laths 30 " 

Roof plastered below rafters or below tie-beam . . . 40 " 
From 75 to 100 feet, add 4 lbs. to each. 

§ 133. Distribution of the Loads. — The methods for de- 
termining the stresses, which will be used here, give correct 
results only when the loads are concentrated at joints, and are 
not distributed over any members of the truss. 

In constructions of importance, this concentration of the 
loads at the joints should always be effected if possible ; 
because, when this is the case, the stresses in the members 
of the truss can be, if properly fitted, obliged to resist only 
stresses of direct tension, or of direct compression ; but, when 
there is a load distributed over any member of the truss, that 
member, in addition to the direct compression or direct tension, 
is subjected to a bending-stress- The effect of this bending 



6 4 



APPLIED MECHANICS. 



cannot be discussed until we have studied the theory of beams. 
Nevertheless, it is a fact that we have no experimental knowl- 
edge of the behavior of a piece under combined compression 
and bending ; and we know that when certain pieces are to 
resist bending, in addition to tension, they must be made much 
larger in proportion than is necessary when they bear tension 
only. 




Fig. 84. 



The manner in which this concentration of the loads is 
effected, is shown in Fig. 84, which is intended to represent one 
of a series of trusses that supports a roof, the rafters being the 
two lower ones in the figure. Resting on two consecutive 
trusses, and extending from one to the other, are beams called 
purlins, which should be placed only above the joints of the truss, 
and which are shown in cross-section in the figure. On these 
purlins are supported the jack-rafters parallel to the rafters, and 
at sufficiently frequent intervals to support suitably the plank 
and superincumbent roofing-materials. 

By this means we secure that the entire bending-stress comes 
upon the jack-rafters and purlins, and that the rafters proper 
are subjected only to a direct compression. When, however, 
the load is distributed, i.e., when the roofing rests directly on the 
rafters, or when the purlins are placed at points other than the 
joints, the bending-stress should be taken into account; and 
while the methods to be developed here will give the stresses 



DIRECT DETERMINATION OF THE STRESSES. 1 65 

in all the members that are not subjected to bending, the bend- 
ing-stress may be largely in excess of the direct stress in those 
pieces that are subjected to bending. How to take this into 
account will be explained later. 

Another important item to consider is, that, in constructions 
of importance, a roller should be placed under one end of the 
truss to allow it to move with the change of temperature ; as 
otherwise some of the members will be either bent, or at least 
subjected to initial stresses. The presence of a roller obliges 
the supporting force at that point to be vertical, whether the 
load be vertical or inclined. 

It is customary, and does not entail any appreciable error, 
to consider the weight of the truss itself, as well as that of the 
superincumbent load, as concentrated at the upper joints ; i.e., 
those on the rafters. 

In the case of a ceiling on the room below, or of a loft 
whose floor rests on the lower joints, we must, of course, ac- 
count the proper load as resting on the lower joints. 

§134. Direct Determination of the Stresses. — This, as 
we have seen, is merely a question of equilibrium of forces in 
a plane, where certain forces acting are known, and others are 
to be determined. 

As to the methods of solution, we might adopt — 

i°. A graphical solution, laying off the loads to scale, and 
constructing the diagram by the use of the propositions of 
the polygon, and the triangle of forces, and then determining the 
results by measuring the lines representing the stresses to 
the same scale. 

2°. An analytical solution, imposing the analytical conditions 
of equilibrium, as given under the " Composition of Forces," 
between the known and unknown forces. 

3 . A third method is to construct the diagram as in the 
graphical solution, but then, instead of determining the results 
by measuring the resulting lines to scale, to compute the un« 



1 66 APPLIED MECHANICS. 

known from the known lines of the diagram by the ordinary 
methods of trigonometry. 

The first, or purely graphical, method, is one which has 
received a very large amount of attention of late years, and 
in which a great deal of progress has been made. It is, doubt- 
less, very convenient for a skilled draughtsman, and especially 
convenient for one who, though skilled in draughting, is not 
free with trigonometric work ; but it seems to me, that, when 
the results are determined by computation from the diagram, 
there is less chance of a slight error in some unfavorable tri- 
angle vitiating all the results. I am therefore disposed to 
recommend for roof-trusses the third method. 

In the case of bridge-trusses, on the other hand, I believe 
the graphical not to be as convenient as a purely analytic 
method. 

§ 135. Roof-Trusses. — In what follows, the graphical solu- 
tions will be explained with very little reference to the trigono- 
metric work, as that varies in each special case, and to one who 
has a reasonable familiarity with the solution of plane triangles, 
it will present no difficulty ; whereas to deduce the formulae 
for each case would complicate matters very much. Proceed- 
ing to special examples, let us take, first, the truss shown in 
Fig. 85, and let the vertical load upon it be W uniformly dis- 
tributed over the top of the roof, the purlins being at the joints 
on the rafters. 

The loads at the several joints will then be as follows, viz. 
(Fig. 85a), - 

ab = kl = — , be = cd = de = ef ' — fg = gh = hk = —-. 
16 8 



Then the supporting forces will be 

lm = ma - 
We thus have, as polygon of external forces, abcdefghklma. 



im — ma = — . 
2 



KOOF-TJ?USSES. 



I6 7 



Now proceed to either support, say, the left-hand one ; and 
there we have the two forces ab and ma known, while by and 
ym are unknown. We then construct 
the quadrilateral maby in the figure, and 
thus determine by and ym. As to whether 





Fig. 85a. 




Fig. 85*5. 



fghM 



abcde 



Fig. 85. 

these represent thrust or tension, 

we need only remember that they 

are the forces exerted by the re- FlG " 8s '' 

spective bars at the joints : and, since by is directed away from 

the bar BY, this bar is in compression; whereas, ym being 

directed towards the bar YM, that bar is in tension. 




1 68 APPLIED MECHANICS. 

Having determined these two stresses, we next proceed to 
another joint, where we have only two unknown forces. Take 
the joint at which the load be acts, and we have as known 
quantities the load be, and also the force exerted by the bar 
YB, which is in compression. This force is now directed away 
from the bar, and hence is represented by yb. The unknown 
forces are the stresses in CX and XY. Hence we construct 
the quadrilateral cxybc ; and we thus determine the stresses in 
CX and XY as ex and xy> both being thrusts. 

Next proceed to the joint YXW, and construct the quadri- 
lateral myxwrn, and thus determine the tension xw and the 
tension wm. 

Next proceed to the joint where cd acts, and so on. We 
thus obtain the diagram (Fig. 8$a) giving all the stresses. 

The truss in the figure was constructed with an angle of 30 s 
at the base, and hence gives special values in accordance with 
that angle. 

In order to show how we may compute the stresses from the 
diagram, the computation will be given. 

From triangle bmy y we have bm = -Z-W 

16 

.-. ym = T-Wcot$o° = ^W 
16 16 

by = -^-^cosec 30 = lw = ky. 
16 8 

From the triangle ume, we have cm = ~ W, ■ 



6 



16 



V*- 



r = \r = '{J; mfi) = 3* 



ROOF-TRUSSES. 



169 



o /V^3 u \ 2 W 

yx — yw sec 30 = ( -^ W\—=. — — = xv = vt t 

\i6 /<J 3 8 
**-*"*-5 Wa= f' " =2 (^) = 8^' 

= \/(^) 2 + (^) 2 = ^\Mr + -^ = ^ W > 
V V 256 256 16 



aw 



f T 2SO 2Kb O 



£# = wm sec 



56 256 
vd = 0*1 sec 30 = /^JfA ~ = $-W 9 

Vie ;v 3 8 

V 4 /v£ 2 



Hence we shall have for the stresses, 



Rafters 


(compression). 


Verticals 


(tension). 


ex = ^!0 




= W> 
= \W. 


##/ = op 


_ W 
16' 


/& = gq 




= %w. 


vu = qr 


_ W 


et =fs 




= i^. 


ts = 


8* 

2 W. 

8 


HORIZONTAL Ties 


(tension) . 










16 


Diagonal Braces (compression). 


my = w« 




xy ■==■ on 


- K 
8 " 


ww = mp 




8 


wv = qp 


JtLW. 

16 


mu ■= mr 




16 


tu = sr 


8 



I/O APPLIED MECHANICS. 

Next, as to the stresses due to wind pressure, we will sup- 
pose that there is a roller under the left-hand end of the truss, 
and none under the right-hand end ; and we will proceed to 
determine the stresses due to wind pressure. 

First, suppose the wind to blow from the left-hand side of 
the truss, and let the total wind pressure be (Fig. 85 £) af = W x . 
The resultant, of course, acts along the dotted line drawn per- 
pendicular to the left-hand rafter at its middle point, as shown 
in Fig. 85. 

The left-hand supporting force will be vertical : hence, pro- 
ducing the above-described dotted line, and a vertical through 
the roller to their intersection, and joining this point with the 
right-hand end of the truss, we have the direction of the right- 
hand supporting force. In this case, since the angle of the 
truss is 30 , the line of action of the right-hand supporting 
force coincides in direction with the right-hand rafter. We 
now construct the triangle of external forces afm t and we 
obtain the supporting forces fm and ma. We then have, as 
the loads at the joints, 

8 y? 

be = -^ = cd= de. 



Then proceed as before to the left-hand joint ; and we find that 
two of the four forces acting there are known, viz., ma and ab, 
and two are unknown, viz., the stresses in BY and YM. Then 
construct the quadrilateral mabym, and we have the stresses by 
and ym ; the first being compression and the second tension, 
as shown by reasoning similar to that previously adopted. 

Then pass to the next joint on the rafter, and construct the 
quadrilateral ybexy, where yb and be are already known, and we 
obtain ex and xy ; and so proceed as before from joint to joint, 



ROOF-TRUSSES WITH LOADS AT LOWER JOINTS. \J\ 



remembering, that, in order to be able to construct the polygon 
of forces in each case, it is necessary that only two of the forces 
acting should be unknown. 

When the wind blows from the other side, we shall obtain 
the diagram shown in Fig. 85^. 

After having determined the stresses from the vertical load 
diagram and those from the two wind diagrams, we should, in 
order to obtain the greatest stress that can come on any one 
member of the truss, add to the stress due to the vertical load 
the greater of the stresses due to the wind pressure. 

§136. Roof-Truss with Loads at Lower Joints. — In 
Fig. 86 is drawn a stress diagram 
for the truss shown in Fig.. 84 on 
the supposition that there is also 
a load on the lower joints. In 
this case let W be the whole load 
of the truss, except the ceiling, 
and W x the weight of the ceiling 
below ; the latter being supported 
at the lower joints and on the 
two extreme vertical suspension 
rods. Then will the loads at the joints be as follows; viz.,— 




Fig. 86. 



ab 


= 


MW+ iv.) 


= kl, 








be 


= 


\{W + IV,) 


= hk, 








cd 


= 


W 


= gh = 


de 


= 


M 


mn 


= 


W* = r 1 


= on = 


qp 


= 


op. 



Observe that there is no joint at the lower end of either of the 
end suspension rods, but that whatever load is supported by 
these is hung directly from the upper joints, where be and hk act. 
We have also for each of the supporting forces Im and ra 



i(W+ JF f ). 



172 APPLIED MECHANICS. 

Hence we have, for the polygon of external forces, 

abcdefghklm nopqra, 

which is all in one straight line, and which laps over on 
itself. 

In constructing the diagram, we then proceed in the same 
way as heretofore. 

§ 137. General Remarks. — As to the course to be pursued 
in general, we may lay down the following directions : — 

1 °. Determine all the external forces ; in other words, the loads 
being known, determine the supporting forces. 

2°. Co7istruct the polygon of forces for each joint of the truss, 
beginning at some joint where only two of the forces acting at 
that joint are unknown. This is usually the case at the support. 
Then proceed from joint to joint, bearing in mind that we can 
only determine the polygon of forces when the magnitudes of 
all but two sides are known. 

3 . Adopt a certain direction of rotation, and adhere to it 
throughout ; i.e., if we proceed in right-handed rotation at 07ie 
joint, we must do the same at all, and we shall thus obtain neat 
and convenient figures. 

4 . Observe that the stresses obtained are the forces exerted 
by the bars under consideration, and that these are thrusts when 
they act away from the bars, and t elisions zvhen they are directed 
towards the bars. 

We will next take some examples of roof-trusses, and con- 
struct the diagrams of some of them, calling attention only to 
special peculiarities in those cases where they exist. 

It will be assumed that the student can make the trigono- 
metric computations from the diagram. 

The scale of load and wind diagram will not always be the 
same ; and the stress diagrams will in general be smaller than 
is advisable in using them, and very much top small if the 



ROOF-TRUSSES WITH LOADS AT LOWER JOINTS. 173 

results were to be obtained by a purely graphical process with- 
out any computation. 

The loads will in all cases be assumed to be distributed 
uniformly over the jack-rafters, or, in other words, concen- 
trated at the joints. 

Those cases where no stress diagram is drawn may be con- 
sidered as problems to be solved. 




Fig. 87. 



Fig. 87a. 




Fig. 87^. 



Fig. 87<t. 



Fig. 87A 



174 



APPLIED MECHANICS. 




Fig. 88. 




Fig. 88a. 




Ygm 




abcdi 



Fig. 88^. 



R O OF- TR USSES WITH LOADS AT LO WER JOINTS. 175 




Fig. 89. 



Fig. 890. 




Fig. 90. Fig. 90a. 



Fig. 91. 



Fig. 91a;. 










/i. 





K 


H 


G 



Fig. 92. 



Fig. 92a. 



Fig. 93. 



Fig. 



176 



APPLIED MECHANICS. 



§138. Hammer- Beam Truss (Fig. 94). — This form of 
truss is frequently used in constructions where architectural 
effect is the principal consideration rather than strength. It 
is not an advantageous form from the point of view of strength, 




Fig. 94A 



Fig. 94c. 



for the absence of a tie-rod joining the two lower joints causes 
a tendency to spread out at the base, which tendency is usually 
counteracted by the horizontal thrust furnished by the but 
tresses against which it is supported. 



HAMMER-BEAM TRUSS. \JJ 



When such a thrust is furnished (or were there a tie-rod), 
and the load is symmetrical and vertical, the bars are not all 
needed, and some of them are left without any stress. In 
the case in hand, it will be found that UV, VM, MQ, and QR 
are not needed. We must also observe that the effect of the 
curved members MY, MV, MQ, and MN on the other parts of 
the truss is just the same as though they were straight, as 
shown in the dotted lines. The curved piece, of course, has to 
be subjected to a bending-stress in order to resist the stress 
acting upon it. If, as is generally the case, the abutments are 
capable of furnishing all the horizontal thrust needed, it will 
first be necessary to ascertain how much they will be called 
upon to furnish. To do this, observe that we have really a truss 
similar to that shown in Fig. 92, supported on two inclined 
framed struts, of which the lines of resistance are the dotted 
lines (Fig. 94) 1 4 and 7 8, and that, under a symmetrical load, 
this polygonal frame will be in equilibrium, and, moreover, the 
curved pieces MV and MQ will be without stress, these only 
being of use to resist unsymmetrical loads, as the snow or 
wind. 

Let the whole load, concentrated by means of the purlins 
at the joints of the rafters, be W. Then will the truss 46 7 have 

W 

to bear \ W, and this will give — to be supported at each of 

4 
the points 4 and 7. Moreover, on the space 2 4 is distributed 

W 

— , which has, as far as overturning the strut is concerned, the 

same effect as — at 2, and — at 4. Hence the load to be sup- 
8 8* F 

ported at 4 by the inclined strut is a vertical load equal to 

(i + i) W — 1 ^ We may then find the force that must be 

furnished by the abutment, or by the tie-rod, in either of the 

two following ways : — 



178 APPLIED MECHANICS 

i°. By constructing the triangle yS € (Fig. 940), with yS == 
I W, ye II 14, and e8 parallel to the horizontal thrust of the abut- 
ment ; then will ySe be the triangle of forces at 1, and e8 will be 
the thrust at 1. 

2 . Multiply |- W by the perpendicular distance from 4 to 
1 2, and divide by the height of 4 above 1 8 for the thrust of the 
abutment ; in other words, take moments about the point 1. 

Now, to construct the diagram of stresses, let, in Fig. 94^, 
the loads be 

ab, be, cd, de, ef, fg, gh, hk, and ki t 
and let 

h = za = \W 

be the vertical component of the supporting force ; let zm be 
the thrust of the abutment : then will Im and ma be the real 
supporting forces ; and we shall have, for polygon of external 
forces, 

abcdefghklma. 

Then, proceeding to the joint 1, we obtain, for polygon of forces, 

maym ; 

and, proceeding from joint to joint, we obtain the stresses in all 
the members of the truss, as shown in Fig. 94^. 

It will be noticed that UV and RQ are also free from 
stress. 

If we had no horizontal thrust from the abutment, and the 
supporting forces were vertical, the members MV and MQ 
would be called into action, and MY and MN'would be inactive. 
To exhibit this case, I have drawn diagram 94^, which shows 
the stresses that would then be developed. A Fand NL would 
become merely part of the supports. 

In this latter case the stresses are generally much greater 
than in the former, and a stress is developed in UV. 



SCISSOR-BEAM TRUSS. 



79 



§ 139. Hammer-Beam Truss: Wind Pressure. — Fig. 95 
shows the stress diagram of the hammer-beam truss for wind 
pressure when there is no roller under either end, and when 
the wind blows from the left. A similar diagram would give the 
stresses when it blows from the right. 




Fig. 95. 



Fig. 95a. 



The cases when there is a roller are not drawn : the student 
may construct them for himself. 

§140. Scissor-Beam Truss. — We have already discussed 
two forms of scissor-beam truss 
in Figs. 90 and 91. These 
trusses having the right number 
of parts, their diagrams present 
no difficulty. Another form of 
the scissor-beam truss is shown 
in Fig. 96, and its diagram pre- 
sents no difficulty. 

The only peculiarity to be noticed is, that, after having con 
structed the polygon of external forces, 

abcdefma, 

we cannot proceed to construct the polygon of equilibrium for 
one of the supports, because there are three unknown forces 




Fig. 96. 



Fig. 96a. 



i8o 



APPLIED MECHANICS. 



there. We therefore begin at the apex CD, and construct the 
triangle of forces cdl for this point ; then proceed to joint CB t 
and construct the quadrilateral 

bclkb; 

then proceed to the left-hand support, and obtain 

mabkgm ; 
and so continue. 

§141. Scissor-Beam Truss without Horizontal Tie. — 

Very often the scissor-beam truss is constructed without any 
horizontal tie, in which case it has the appearance of Fig. 97, 
where there is sometimes a pin at GKLH and sometimes not. 





Fig. 97a. 




■*7 



*k7 

I 



Fig. 97. 



Fig. 973. 



Fig. 97c. 



In this case, if the abutments are capable of furnishing hori* 
zontal thrust to take the place of the horizontal tie of Fig. 96, 
we are reduced back to that case. If the abutments are not 
capable of furnishing horizontal thrust, we are then relying on 
the stiffness of the rafters to prevent the deformation of the 
truss ; for, were the points BC and DE really joints, with pins, 
the deformation would take place, as shown in Fig. 97^ or Fig. 
97$, according as the two inclined ties were each made in one 
piece or in two (i.e., according as they are not pinned together 
at KH, or as they are pinned). This necessity of depending 
on the stiffness of the rafters, and the liability to deformation 
if they had joints at their middle points, become apparent as 
soon as we attempt to draw the diagram. Such an attempt is 



SCISSOR-BEAM TRUSS WITHOUT HORIZONTAL TIE. l8l 

made in Fig. gye, where abcdefga is the polygon of external 
forces, gabkg the polygon of stresses for the left-hand support, 
kbclk that for joint BC. Then, on proceeding to draw the tri- 
angle of stresses for the vertex, we find that the line joining d 
and / is not parallel to DL, and hence that the truss is not 
stable. We ought, however, in this latter case, when the sup- 
porting forces are vertical, and when we rely upon the stiffness 
of the rafters to prevent deformation, to be able to determine 
the direct stresses in the bars ; and for this we will employ an 
analytical instead of a graphical method, as being the most con- 
venient in this case. 

Let us assume that there is no pin at the intersection of the 
two ties, and that the two rafters are inclined at an angle of 45 ° 
to the horizon. 

We then have, if W = the entire load, and a = angle 
between BK and KG, 

ab = ef = — , be = cd = de = — , 
8' 4 



"5"? 



I 2 

Sin a = -— , COS a = 



Let x be the stress in each tie, and let y = cl = dl = thrust 
in each upper half of the rafters. 

Then we must observe that the rafter has, in addition to its 
direct stresses, a tendency to bend, due to a normal load at the 
middle, this normal load being equal to the sum of the normal 
components of be and of x, when these are resolved along and 
normal to the rafter. Hence 

normal load = ^cosa -\ sin 45 °. 

4 

This, resolved into components acting at each end of the rafter, 
gives a normal downward force at each end equal to 

£#cosa -f- -J^sin45°. 



152 APPLIED MECHANICS. 

Hence, resolving all the forces acting at the left-hand support 
into components along and at right angles to the rafter, and 
imposing the condition of equilibrium that the algebraic sum 
of their normal components shall equal zero, we have, if we call 
upward forces positive, 

f fFsin45° — (-Jxcosa + ^-£Fsin45°) — #sina. = o; (i) 

but, since 

#cosa = 2x sin a, 
we have from (i) 

2X sin a = — sin 45 
4 

W . 
.'. jfsma = — sin 45 

... x = i^!iM3! (2) 

sma 

Then, proceeding to the apex of the roof, we have that the load 

cd — — 
4 

gives, when resolved along the two rafters, a stress in each 
equal to 

W . o 
— sin 45 . 

4 ** 

Hence the load to be supported in a direction normal to the 

rafter at the apex is 

— sin 45 -f- (-J* cos a + —sin 45 ). 
4 o 

Hence, substituting for x its value, we have 

y = cl'= dl = —sin 45 . (3) 

2 

Then, proceeding to the left-hand support, and equating to zero 
the algebraic sum of the components along the rafter, we have 

bk = (ga — ab) cos 45 + ^cosa 

= f ^sin 45° + i ^sin 45° = | ^sin 45°. (4) 



SCISSOR-BEAM TRUSS WITHOUT HORIZONTAL TIE. 1 83 



We have thus determined in (2), (3), and (4) the values of x, y y 
and bk = eh. 

By way of verification, proceed to the middle of the left- 
hand rafter, and we find the algebraic sum of the components 
of be and x along the rafter to be 

JJ^cos45° ~ -*sina = \ W€ra^ \ 

and this is the difference between bk and el, as it should be. 

We have thus obtained the direct stresses ; and we have, in 
addition, that the rafter itself is also subjected to a bending- 
moment from a normal load at the centre, this load being equal 

to 

W o W 

.* cos a H sin 45 = — sin 45 . 

4 2 

How to take this into account will be explained under the 
" Theory of Beams." 

§142. Examples. — The following figures of roof-trusses 
may be considered as a set of examples, for which the stress 
diagrams are to be worked out. 

Observe, that, wherever there is a joint, the truss is to be 
supposed perfectly flexible, i.e., free to turn around a pin. 



Fig. 98. 



Fig. 99. 



Fig. 100. 



Fig. i 01. 




Fig. 102. 




Fig. 103. 




Fig. 106. 



Fig. 



Fig. 104. 





Fig. iov 



Fig. 108 



1 84 APPLIED MECHANICS. 



CHAPTER IV. 
BRIDGE-TRUSSES. 

§143. Method of Sections. — It is perfectly possible to 
determine the stresses in the members of a bridge-truss 
graphically, or by any methods that are used for roof-trusses. 

In this work an analytical method will be used ; i.e., a method 
of sections. This method involves the use of the analytical con- 
ditions of equilibrium for forces in a plane explained in § 63. 
These are as follows ; viz., — 

If a set of forces in a plane, which are in equilibrium, be 
resolved into components in two directions at right angles to 
each other, then — 

i°. The algebraic sum of the components in one of these 
directions must be zero. 

2 . The algebraic sum of the components in the other of 
these directions must be zero. 

3 . The algebraic sum of the moments of the forces about 
any axis perpendicular to the plane of the forces must be zero. 

Assume, now, a bridge-truss (Figs. 109, no, in, 112, pages 
j 86 and 187) loaded at a part or all of the joints. Conceive a 
vertical section ab cutting the horizontal members 6-8 and 7-9 
and the diagonal y-8, and dividing the truss into two parts. 
Then the forces acting on either part must be in equilibrium,, 
in other words, the external forces, loads, and supporting forces,, 
acting on one part, must be balanced by the stresses in the 
members cut by the section ; i.e., by the forces exerted by the 
other part of the truss on the part under consideration. Hence 
we must have the three following conditions ; viz., — 



SHEARIXG-FORCE AND B ENDING-MOMENT. 185 

i°. The algebraic sum of the vertical components of the 
above-mentioned forces must be zero, 

2°. The algebraic sum of the horizontal components of these 
forces must be zero. 

3 . The algebraic sum of the moments of these forces about 
any axis perpendicular to the plane of the truss must be zero. 

§ 144. Shearing-Force and Bending-Moment. — Assum- 
ing all the loads and supporting forces to be vertical, we shall 
have the following as definitions. 

The Shearing-Force at any section is the force with which 
the part of the girder on one side of the section tends to slide 
by the part on the other side. 

In a girder free at one end, it is equal to the sum of the 
loads between the section and the free end. 

In a girder supported at both ends, it is equal in magnitude 
to the difference between the supporting force at either end, 
and the sum of the loads between the section and that support- 
ing force. 

The Bending-Moment at any section is the resultant moment 
of the external forces acting on the part of the girder to one side 
of the section, tending to rotate that part of the girder around 
a horizontal axis lying in the plane of the section. 

In a girder free at one end, it is equal to the sum of the 
moments of the loads between the section and the free end, 
about a horizontal axis in the section. 

In a girder supported at both ends, it is the difference be- 
tween the moment of either supporting force, and the sum of 
the moments of the loads between the section and that sup- 
port ; all the moments being taken about a horizontal axis in 
the section. 

§ 145. Use of Shearing-Force and Bending-Moment. — ■ 
The three conditions stated in § 143 may be expressed as fol- 
lows : — 

i°. The algebraic sum of the horizontal components of the 
stresses in the members cut by the section must be zero. 



1 86 



APPLIED MECHANICS. 



2°. The algebraic sum of the vertical components of the 
stresses in the members cut by the section must balance the 
shearing-force. 

3°. The algebraic sum of the moments of the stresses in 
the members cut by the section, about any axis perpendicular to 
the plane of the truss, and lying in the plane of the section, 
must balance the bending-moment at the section. 

As the conditions of equilibrium are three in number, they 
will enable us to determine the stresses in the members, pro- 
vided the section does not cut more than three ; and this 
determination will require the solution of three simultaneous 
equations of the first degree with three unknown quantities 
(the stresses in the three members). 

By a little care, however, in choosing the section, we can 
very much simplify the operations, and reduce our work to the 
solution of one equation with only one unknown quantity ; the 
proper choice of the section taking the place of the elimination. 

§146. Examples of Bridge-Trusses. — Figs. 109-1 12 rep- 
resent two common kinds of bridge-trusses : in the first two 

the braces are all 

x 11 13 15 17 19 21 23 25 27 29 



v\/v v 



4 6 & 



7 WW\AAAAA4 



10 12 14 16 18 20 22 24 26 28 



diagonal, in the 
last two they are 
partly vertical and 
FlG - Iog - partly diagonal. 

The first two are called Warren girders, or half-lattice girders ; 
since there is only one system of bracing, 
as in the figures. When, on the other 
hand, there are more than one system, so 
that the diagonals cross each other, they 
are called lattice girders. 

§ 147. General Outline of the Steps 
to be taken in determining the Stresses 
in a Bridge-Truss under a Fixed Load. 

i°. If the truss is supported at both ends, find the sup- 
porting forces. 




Fig. i 10. 



DETERMINING THE STRESSES IN A BRIDGE-TRUSS. 1 87 



2°. Assume, in all cases, a section, in such a manner as not 
to cut more than three members if possible, or, rather, three 
of those that 

11 13 15 17 19 21 23 25 27 28 



la 



10 12 14 16 18 20 22 24 26 



11 13 



Fig. 112. 



are brought 
into action 
by the loads 
on the truss ; 

and it will FlG - "*• 

save labor if we assume the section so as to cut two of the 

three very near their point of inter- 
section. 

3 . Find the shearing-force at the 
section. 

4 . Find the bending-moment at 
the section. 

5°. Impose the analytical conditions of equilibrium on all 
the forces acting on the part of the girder to one side of the 
section, — the part between the section and the free end when 
the girder is free at one end, or either part when it is supported 
at both ends. 

In the cases shown in Figs. 109 and no, we may describe 
the process as follows ; viz., — ■ 

(a) Find the stress in the diagonal from the fact, that (since 
the stress in the diagonal is the only one that has a vertical 
component at the section) the vertical component of the stress 
in the diagonal must balance the shearing-force. 

(b) Take moments about the point of intersection of the 
diagonal and horizontal chord near which the section is taken ; 
then the stresses in those members will have no moment, so 
that the moment of the stress in the other horizontal must 
balance the bending-moment at the section. Hence the stress 
in the horizontal will be found by dividing the bending-moment 
at the section by the height of the girder. 

The above will be best illustrated by some examples. 



1 88 APPLIED MECHANICS. 

Example I. — Given the semi-girder shown in Fig. no, 
loaded at joint 13 with 4000 pounds, and at each of the joints 
i> 3> 5> 7> 9> an d 11 with 8000 pounds. Suppose the length of 
each chord and each diagonal to be 5 feet. Required the stress 
in each member. 

Solution. — For the purpose of explaining the method of 
procedure, we will suppose that we desire to find first the 
stresses in 8-10 and 9-10. 

Assume a vertical section very near the joint 9, but to the 
right of it, so that it shall cut both 8-10 and 9-10. 

If, now, the truss were actually separated into two parts at 
this section, the right-hand part would, in consequence of the 
loads acting on it, separate from the other part. This tendency 
to separate is counteracted by the following three forces : — 

i°. The pull exerted by the part 9-x of the bar 9- n on the 
part x-11 of the same bar. 

2 . The thrust exerted by the part 8-z of the bar 8-10 on 
the part £-10 of the same bar. 

3 . The pull exerted by the part 9-7 of the bar 9-10 on the 
part y- 10 of the same bar. 

The shearing-force at this section is 

8000 4- 4000 = 1 2000 lbs., 

and this is equal to the vertical component of the stress in the 
diagonal. Hence 

Stress in 9-10 == = 12000(1.1547) = 13856 lbs. 

sin 60 

This stress is a pull, as may be seen from the fact, that, in 
order to prevent the part of the girder to the right of the 
section from sliding downwards under the action of the load, 
the part g-y of the diagonal 9-10 must pull the part y-\o of 
the same diagonal. 

Next take moments about 9 : and, since the moment of the 
stresses in 9-1 1 and 9-10 about 9 is zero, we must have that the 
moment of the stress in 8-10; i.e., the product of this stress 
by the height of the girder, must equal the bending-moment. 



DETERMINING THE STRESSES IN A BRIDGE-TRUSS. 1 89 



The bending-moment about 9 is 

8000 X 5 4- 4000 x 10 = 80000 foot-lbs. 
80000 



Hence 



Stress in 8-10 == 



4-33 



80000(0.23094) = 18475 1 DS » 



Proceed in a similar way for all the other members. The 
work may be arranged as in the following table ; the diagonal 
stresses being deduced from the shearing-forces by multiplying 
by 1. 1 547, and the chord stresses from the bending-moments 
by multiplying by 0.23094. 




.2. .2 

'C 
« J* 


Shearing- 
Force 
in lbs. 


Stresses in Diagonals cut 
by Section, in lbs. 


Bending- 

Moment, in 

foot-lbs. 


Stresses in Chords opposite the 
respective Joints. 


Tension. 


Compression. 


Tension. 


Compression. 


I 


44OOO 


50806 




72OOOO 




166277 


2 


44OOO 




50806 


6lOOOO 


I40873 




3 


360OO 


41569 




5OOOOO 




11547° 


4 


360OO 




41569 


4IOOOO 


94685 




5 


280OO 


3 2 33I 


\ 32OOOO 




739OI 


6 


280OO 




3 2 33* | 250000 


57735 




7 


20000 


23O94 


i 180000 




41569 


8 


20000 




23094 


I3OOOO 


30022 




9 


12000 


13856 




80O00 




18475 


10 


I2000 




13856 


5OOOO 


"547 




11 


4OOO 


4619 




20000 




4618 


12 


4OOO 




4619 


I OOOO 


2309 





Example II. — Given the truss (Fig. 109) loaded at each ol 
the lower joints with 10000 lbs. : find the stresses in the members. 
The length of chord is equal to the length of diagonal = 10 ft. 

Throughout this chapter, tensions will be written with the 
minus, and compressions with the plus sign. 

Solution. — Total load = 14(10000) = 140000 lbs. 

Each supporting force = 70000 " 
The entire work is shown in the following tables: — 



1 90 



APPLIED MECHANICS. 









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DETERMINING THE STRESSES IN A BRIDGE-TRUSS. 191 



Numbers 


of Diagonals. 


Stresses 


in Diagonals, in lbs. 


I- 2 


28-29 


— 70000 x 


1. 1547 = —80829 


2- 3 


27-28 


+ 60000 x 


1. 1547 = +69282 


3" 4 


26-27 


— 60000 x 


1. 1547 = —69282 


4- 5 


25-26 


+ 50000 x 


1. 1547 = +57735 


5" 6 


24-25 


— 50000 x 


1. 1547 = -57735 


6- 7 


' 23-24 


+40000 x 


1. 1547 = +46188 


7-8 


22-23 


— 40000 x 


1. 1547 = —46188 


8-9 


21-22 


+ 30000 X 


i-i547 = +34641 


9-10 


20-2I 


— 30000 x 


i-i547 = -34641 


10-11 


I9-2O 


+ 20000 X 


i-i547 = +23094 


11-12 


18-I9 


— 20000 x 


1. 1547 = -23094 


12-13 


I7-l8 


+ IOOOO x 


i-i547 = +II547 


13-M 


16-17 


— 10000 X 


1. 1547 = -11547 


14-15 


I5-I6 


+0 






LOWER CHORDS. 



lumbers of Chords. 


Stresses in Chords, in 


lbs. 


2- 4 


26-28 


— 650000 X 0.1 1547 = 


- 75056 


4- 6 


24-26 


— 1200000 X 0.11547 = 


-I38564 


6- 8 


22-24 


— 1650000 X O.11547 = 


— I9O526 


8-10 


20-22 


— 2000000 X 0.1 1547 = 


— 23O94O 


10-12 


18-20 


— 2250000 X 0.1 1547 = 


— 259808 


12-14 


16-18 


— 2450000 X O.I 1547 = 


— 277128 


14-16 




— 2450000 X 0.1 1547 = 


— 2829O2 



9 2 



APPLIED MECHANICS. 



UPPER CHORDS. 


Numbers of Chords. 


Stresses in Chords, in lbs. 


*'- 3 


27-29 


350000 x 0.1 1547 = + 40415 


3- 5 


25-27 


950000 X 0.11547 = +109697 


5" 7 


2 3-25 


1450000 X 0.1 1547 = +167432 


7- 9 


21-23 


1850000 X 0.11547 = +213620 


9-1 1 


19-21 


2150000 X 0.11547 = +248261 


11--13. 


17-19 


2350000 X 0.11547 = +267355 


•3-15 


15-17 


2450000 X O.I 1547 = +282902 



Example III. — Given the same truss as in Example II., 
loaded at 2, 4, 6, 8, 10, and 12 with 10000 lbs. at each point, 
the remaining lower joints being loaded with 50000 lbs. at each 
joint : find the stresses in the members. 

Example IV. — Given a semi-girder, free at one end (Fig. 
112), loaded at 2, 4, and 6 with 10000 lbs., and at 8, 10, and 12 
with 5000 lbs. : find the stresses in the members. 



TRAVELLING-LOAD. 

§148. Half- Lattice Girder: Travelling-Load. — When a 
girder is used for a bridge, it is not subjected all the time to 
the same set of loads. 

The load in this case consists of two parts, — one, the dead 
load, including the bridge weight, together with any permanent 
load that may rest upon the bridge ; and the other, the moving 
or variable load, also called the travelling-load, such as the 
weight of the whole or part of a railroad train if it is a railroad 
bridge, or the weight of the passing teams, etc., if it is a common- 
road bridge. Hence it is necessary that we should be able to 
determine the amount and distribution of the loads upon the 
bridge which will produce the greatest tension or the greatest 



GREATEST DIAGONAL STRESSES IN GIRDER. T93 

compression in every member, and the consequent stress pro- 
duced. 

§149. Greatest Stresses in Semi-Girder. — Wherever the 
section be assumed in a semi-girder, it is evident that any load 
placed on the truss at any point between the section and the 
free end increases both the shearing-force and the bending- 
moment at that section, and that any load placed between the 
section and the fixed end has no effect whatever on either 
the shearing-force or the bending-moment at that section. 

Hence every member of a semi-girder will have a greater 
stress upon it when the entire load is on, than with any partial 
load. 

§ 150. Greatest Chord Stresses in Girder supported at 

Both Ends Every load which is placed upon the truss, no 

matter where it is placed, will produce at any section whatever a 
bending-moment tending to turn the two parts of the truss on 
the two sides of the section upwards from the supports ; i.e., so 
as to render the truss concave upwards. 

Hence every load that is placed upon the truss causes com- 
pression in every horizontal upper chord, and tension in every 
horizontal lower chord. Hence, in order to obtain the greatest 
chord stresses, we assume the whole of the moving load to be 
upon the bridge. 

§ 151. Greatest Diagonal Stresses in Girder supported 
at Both Ends. — To determine the distribution of the load 
that will produce the greatest stress of a certain kind (tension 
or compression) in any given diagonal, let us suppose the diag- 
onal in question to be 7-8 (Fig. 109), through which we take 
our section ab. Now it is evident that any load placed on the 
truss between ab and the left-hand (nearer) support will cause a 
shearing-force at that section which will tend to slide the part 
of the girder to the left of the section downwards with refer- 
ence to the other part, and hence will cause a compressive 
stress in 7-8 ; while any load between the section and the right- 



194 APPLIED MECHANICS. 

hand (farther) support will cause a shearing-force of the oppo- 
site kind, and hence a tension in the bar y-S. 

Now, the bridge weight itself brings an equal load upon each 
joint ; hence, when the bridge weight is the only load upon the 
truss, the bar y-8 is in tension. 

Hence, any load placed upon the truss between the section 
and the farther support tends to increase the shearing-force at 
that section due to the dead load (provided this is equally dis- 
tributed among the joints) ; whereas any load placed between 
the section and the nearer support tends to decrease the shear- 
ing-force at the section due to the dead load, or to produce a 
shearing-force of the opposite kind to that produced by the dead 
load at that section. 

Hence, if we assume the dead load to be equally distributed 
among the joints, we shall have the two following propositions 
true : — 

(a) In order to determine the greatest stress in any diagonal 
which is of the same kind as that produced by the dead load, 
we must assume the moving load to cover all the panel points 
between the section and the farther abutment, and no other 
panel points. 

(b) In order to determine the greatest stress in any diagonal 
of the opposite kind to that produced by the dead load, we must 
assume the moving load to cover all the panel points between 
the section and the nearer abutment, and no others. 

This will be made clear by an example. 

Example I. — Given the truss shown in Fig. 113. Length 

of chord = length of diagonal =. 



^\ AAAAAAA %^ I0 feet ' Dead load = 8oo ° lbs " 

' applied at each upper panel point. 

Moving load =± 30000 lbs. applied 

at each upper panel point. Find 



8 10 12 14 

Fig. 113. 



the greatest stresses in the members. 



EXAMPLE OF BRIDGE-TRUSS. 



195 



Solution, (a) Chord Stresses. — Assume the whole load to 
be upon the bridge : 



this will give 



+. 208423 (5) + 296181 (7; + 340059 (9) 




38OOO 

lbs. at "" each upper 
panel point ; i.e., omit- 
ting 1 and 17, where 
the load acts directly 
on the support, and 
not on the truss. 
Hence, considering the bridge so loaded, we shall have the fol- 
lowing results for the chord stresses : — 

Each supporting force = 38ooo(-J = 133000. 



Fig. 114. 



Section at 


Bending-Moment, in foot-lbs. 


2 


16 


133000 X 


5 


= 665000 


3 


*5 


133000 x 


10 


= 1330000 


4 


14 


133000 x 


15 


— 38000 X 5 = 1805000 


5 


13 


133000 x 


20 


— 38000 X 10 = 2280000 


6 


12 


133000 x 


2 5 


- 3 8ooo ( 5 + 15) = 2 5 6 5°oo 


7 


11 


133000 x 


30 


— 38000(10 -J- 20) = 2850000 


8 


10 


133000 x 


35 


- 38ooo( 5 + 15 + 25) = 2945000 


9 




133000 x 


40 


— 38000(10 4- 20 + 30) = 3040000 



Numbers of Chords. 


Stresses in Upper Chords. 


i-3 

3-5 
5-7 
7-9 


i5- J 7 

^3-i5 

11-13 

9-1 1 


665000 X O.11547 = + 76788 
1805000 X 0.1 1547 = +208423 
2565000 X 0.11547 = +296181 
2945000 X 0.1 1547 = +340059 



9 6 



APPLIED MECHANICS. 



Numbers of Chords. 


Stresses in Lower Chords. 


2- 4 
4- 6 
6- 8 
8-io 


14-16 
12-14 
10-12 


-1330000 x 0.11547 = -153575 

— 2280000 X 0.11547 = —263272 

— 2850000 X O.I 1547 = —329090 

— 3040000 X 0.11547 = -351029 



Next, as to the diagonals, take, for instance, the diagonal 
7-8. When the dead load alone is on the bridge, the diagonal 
7-8 is in tension. From the preceding, we see that the greatest 
tension is produced in this bar when the moving load is on the 
points 9, 11, 13, and 15, and the dead load only on the points 3, 
5, 7. Now, a load of 38000 lbs. at 13, for instance, causes a 

shearing-force of -^(38000) = 9500 lbs. at any section to the 
16 

left of 1 3 ; and this shearing-force tends to cause the part to 

the left of the section to slide upwards, and that to the right 

downwards. 

On the other hand, with the same load at the same place, 

1 2 
there is produced a shearing-force of — (38000) = 28500 lbs. 

at any section to the right of 13 ; and this shearing-force tends 
to cause the part to the left to slide downwards, and that to the 
right upwards. Paying attention to this fact, we shall have, 
when the loads are distributed as above described, a shearing- 
force at the bar y-8 causing tension in this bar ;. the magnitude 
of this shearing-force being 

3^2( 2 + 4 + 6 + 8) - ^( 2 + 4 + 6) = 4iS°o. 

l6 l6 



Hence, we may arrange the work as follows : 



GREATEST DIAGONAL STRESSES IN GIRDER. 



I97 







Greatest 






Stresses in 


Numbers 0/ 
Diagonals. 


Greatest Shearing- Forces producing Stresses of Same Kind as 
Dead Load. 


Diagonals of 

Same Kind 

as those due 

to Dead 






Lead. 


1-2 


17-16 


^-^(2 + 4+6+8+10+12 + 14) = 133000 


-153575 


2-3 


16-15 


^-^(2 + 4+6+8+10+12+14) = 1330OO 


+ 153575 


3-4 


15-H 


3^ (2 + 4 + 6+8+IO+I2) -^(2) = 98750 


— 1 14027 


4-5 


14-13 


^(2+4+6+8+10+12) -^(2) = 98750 


+ 1 1+027 


5-6 


13-12 


^(2+4+6+8+10) -^(2+4) = 68250 


— 78808 


6-7 


1 2-1 1 


^(2+4+6+8+10) -^(2+4) = 68250 


+ 78808 


7-8 


II-IO 


^(2+4+6+8) - ^(2+4+6) = 41500 


— 47920 


8-9 


10- 9 


^(2+4+6+8) - ^(2+4+6) = 41500 


+ 479 2 







Greatest 






Stresses in 


Numbers of 


Greatest Shearing-Forces producing Stresses of Kind Opposite 


Diagonals of 


Diagonals. 


from Dead Load. 


Kind Oppo- 
site from 
Dead Load. 


8-9 


10- 9 


^P<2+ 4 +6) - —(2+4+6+8) = 18500 


— 21362 


7-8 


II-IO 


^(2+4+6) - §=? (2+4+6 +8) = .8500 


+ 2 1 362 



The diagonals 7-8, 8-9, 9-10, and 10-11 are the only ones 
that, under any circumstances, can have a stress of the kind 
opposite to that to which they are subjected under the dead 
load alone. 



I9 8 APPLIED MECHANICS. 

Fig. 114 exhibits the manner of writing the stresses on the 
diagram. 

§152. General Application of this Method. — It is plain 
that the method used above will apply to any single system of 
bridge-truss with horizontal chords and diagonal bracing, what- 
ever be the inclination of the braces. 

When seeking the stress in a diagonal, the section must be 
so taken as to cut that diagonal ; and, as far as this stress alone 
is concerned, it may be equally well taken at any point, as well 
as near a joint, provided only it cuts that diagonal which is in 
action under the load that produces the greatest stress in this 
one, and no other. 

On the other hand, when we seek the stress in a horizontal 
chord, the section might very properly be taken through the 
joint opposite that chord. 

Taking it very near the joint, only serves to make one sec- 
tion answer both purposes simultaneously. 

§ 153. Bridge-Trusses with Vertical and Diagonal Bra- 
cing. — When, as in Figs, in and 112, there are both vertical 
and diagonal braces, and also horizontal chords, we may deter- 
mine the stresses in the diagonals and in the chords just as 
before ; only we must take the section just to one side of a joint, 
and never through the joint. 

As to the verticals, in order to determine the stress in any 
vertical, we must impose the conditions of equilibrium between 
the vertical components of the forces acting at one end of that 
vertical : thus, if the loads are at the upper joints in. Fig. in, 
then the stress in vertical 3-2 must be equal and opposite to 
the vertical component of the stress in diagonal 1-2, as these 
stresses are the only vertical forces acting at joint 2. 

Vertical 5-4 has for its stress the vertical component of the 
stress in 3-4, etc. Thus 

Stress in 3-2 = shearing-force in panel 1-3, 
Stress in 5-4 = shearing- force in panel 3-5, ete. 



TRUSSES WITH VERTICAL AND DIAGONAL BRACING. 1 99 

On the other hand, if the loads be applied at the lower 
joints, then 

Stress in 3-2 = shearing-force in panel 3-5, 
Strgss in 5-4 = shearing-force in panel 5-7, etc. 

Example. — Given the truss shown in Fig. in. Given 
panel length = height of truss = 10 feet, dead load per panel 
point = 12000 lbs., moving load per panel point — 23000 lbs. ; 
load applied at upper joints. 

Solution, (a) Chord Stresses. — Assume the entire load on 
the bridge, i.e., 35000 lbs. per panel point. Hence 

Total load on truss =13 (35000) = 455000 lbs., 
Each supporting force = 227500 lbs. 



Joint near 






which 
Section is 


Bending-Moment at the Section very near the Joint, on Either Side of the Joint. 


taken. 






I 


28 


O 




3 


27 


227500 X IO 


= 2275000 


5 


25 


227500 X 20 — 35000 X 10 


= 4200000 


7 


23 


227500 x 30 — 35000(10 4- 20) 


= 5775000 


9 


21 


2275OO X 40 — 350OO (IO + 20 + 30) 


= 7000000 


n 


19 


227500 x 50 — 35000 (10 -f 20 + 30 + 40) 


= 7875000 


13 


17 


227500 x 60 — 35000(10 + 20 4- 30 + 40 + 50) 


= 8400000 


15 


— 


227500 x 70 — 35000(10 -|- 20 +■ 30 4- 40 4- 50 4- 60) 


= 8575000 



To find any chord stress, divide the bending-moment at a 
section cutting the chord, and passing close to the opposite 
joint, by the height of the girder, which in this case is 10. 
Hence we have for the chord stresses (denoting, as before, com- 
pression by -f-, and tension by — ) : — 



20O- 



APPLIED MECHANICS. 



Stresses in Upper Chords. 


Stresses in Lower Chords. 


i- 3 


27-28 


+ 227500 


2- 4 


24-26 • 


— 227500 


3- 5 


25-27 


+420000 


4- 6 


22-24 


— 420000 


5- 7 


2 3- 2 5 


+ 5775°° 


6- 8 


20-22 


-577500 


7" 9 


21-23 


+ 700000 


8-10 


18-20 


— 700000 


9-1 1 


19-21 


+ 787500 


10-12 


16-18 


-787500 


11-13 


17-19 


+ 840000 


12-14 


14-16 


— 840000 


13-15 


i5-!7 


+ 857500 









Diagonals. — It is evident, that, for the diagonals, the same 
rule holds as in the case of the Warren girder : i.e., the greatest 
stress of the same kind as that produced by the dead load 
occurs when the moving load is on all the joints between the 
diagonal in question and the farther abutment ; whereas the 
greatest stress of the opposite kind occurs when the moving 
load covers all the joints between the diagonal in question and 
the nearer abutment. 

The work of determining the greatest shearing-forces may 
be arranged as in tables on p. 191. 

Counterbraces. — If the truss were constructed with those 
diagonals only that slope downwards towards the centre, and 
which may be called the main braces, the diagonals n-12, 
13-14, 14-17, and 16-19 would sometimes be called upon to 
bear a thrust, and the verticals 12-13 and 17-16 a pull : this 
would necessitate making these diagonals sufficiently strong 
to resist the greatest thrust to which they are liable, and fixing 
the verticals in such a way as to enable them to bear a pull. 

In order to avoid this, the diagonals 10-13, 12-15, I 5~ I 6, 
and 17-18 are inserted, which are called counterbraces, and 
which come into action only when the corresponding main 



TRUSSES WITH VERTICAL AND DIAGONAL BRACING. 20I 

braces would otherwise be subjected to thrust. They also 
prevent any tension in the verticals. 



Diagonals. 


Greatest Shearing-Forces of the Same Kind as those produced by 
Dead Load. 


I- 2 


28-26 


3 -^(i+2+3+...+i3) =227500 


3- 4 


27-24 


*=Pfi + »+3+.., + M)-=PM =194M3 


5-6 


25-22 


^, + a + 3 +... + ll)- 1 -=={l + 2) =162429 


7-8 


23-20 


22f(i4-2+34-...4-io)-'-f?(i + 24-3) =132357 


9-10 


21-18 


2^(1 + 2 + 3+...+ 9 )-^?(i + 2+...+4)= 103929 


11-12 


19-16 


^(14-2+3+...+ 8)-i^?(i + 2+...+ 5 )= 77H3 


13-14 


17-14 


^(,4-24-34-... .+ 7>- Ij =ptH-2+...+6)= 52000 



Diagonals. 


Greatest Shearing-Forces of the Opposite Kind to 
Dead Load. 


those produced by 


I3-H 


17-14 


H», +a+s+ ...+ 6)- =£fr+a+:. 

14 14 


.+7)= 28500 


II-I2 


I9-16 


2^?(i + 2+ ... + 5)-^(. + 2+.. 


.+8)= 6643 


9-IO 


2I-I8 


«=*I + 3 + ... + 4)-^(l + 2+.. 


• + 9) = — i357i 



The main braces and counterbraces of a panel are never in 
action simultaneously. Hence we have, for the greatest stresses 
in the diagonals, the following results, obtained by multiplying 



the corresponding shearing-forces by 



cos45 ( 



= 1.414. 



202 



APPLIED MECHANICS. 



In the following I have used this number to three decimal 
places, as being sufficiently accurate for practical purposes. 



Stresses in Mair 


l Braces. 


Stresses in Counterbraces. 


I- 2 


28-26 


-3 2l68 5 


15-12 


15-16 


— 40299 


3- 4 


27-24 


-274518 


I3-IO 


I7-I8 


- 9393 


5" 6 


25-22 


— 229675 








7- 8 


23-20 


-187153 








9-10 


21-18 


— 146956 








11-12 


19-16 


— 109080 








13-H 


17-14 


- 735 2 8 









Vertical Posts. — Since the loads are applied at the upper 
joints, the conditions of equilibrium at the lower joints require 
that the thrust in any vertical post shall be equal to the vertical 
component of the tension in that diagonal which, being in action 
at the time, meets it at its lower end. 

Hence it is equal to the shearing-force in that panel where 
the acting diagonal meets it at its lower end. 

We therefore have, for the posts, the following as the greatest 
thrusts: — ■ 

STRESSES IN VERTICALS. 



3- 2 


27-26 


+ 227500 


5- 4 


25-24 


+ 194143 


7- 6 


23-22 


+ 162429 


9- 8 


21-20 


+ x 3 2 357 


II-IO 


19-18 


+ 103929 


13-12 


17-16 


+ 77143 


15- 


-14 


+ 52000 



CONCENTRATING THE LOAD AT THE JOINTS. 



203 



1 





Fig. 115 shows the stresses marked on the diagram. 

§ 154. Manner of Concentrating the Load at the Joints. 

— In using the methods given above, we are j- 

assuming that all the loads are concentrated 

at the joints, and that none are distributed 

over any of the pieces. As far as the mov- 
ing load is concerned, and also all of the 

dead load except the weight of the truss 

itself, this always is, or ought to be, effected ; 

and it is accomplished in a manner similar 

to that adopted in the case of roof-trusses. 

This method is shown in the figure (Fig. 

116); floor-beams being laid across from 
girder to girder at the joints, 
on top of which are laid longi- 
tudinal beams, and on these 
the sleepers if it is a railroad 
bridge, or the floor if it is a 
road bridge. The weight of 
the truss itself is so small a 
part of what the bridge is 
called upon to bear, that it 
can, without appreciable error, 
be considered as concentrated 
at the joints either of the up- 
per chord, of the lower chord, 
or of both, according to the 
manner in which the rest of 
the load is distributed. 

§ 155. Closer Approxima- 
tion to Actual Shearing- 
Force. — In our computations 
of greatest shearing-force, we 






Fig. 115. 



make an approximation which is generally considered to be 



20 4 APPLIED MECHANICS. 

sufficiently close, and which is always on the safe side. To 
illustrate it, take the; case of panel 3-5 of the last example. 
In determining its greatest shearing-force, we considered a load 
of 35000 lbs. per panel point to rest on all the joints from the 
right-hand support to joint 5, inclusive, and the dead load to 
rest on all the other joints of the truss. Now, it is impossible, 
if the load is distributed uniformly on the floor of the bridge, 
to have a load of 35000 lbs. on 5 and 12000 on 3 simultaneously ; 
for, if the moving load extended on the bridge floor only up to 
5, the load on 5 would be only 12000 + ^(23000) = 23500 lbs., 
and that on 3 would then be 12000 lbs. If, on the other hand, 
the moving load extends beyond 5 at all, as it must if the load 
on 5 is to be greater than 23500 lbs., then part of it will rest 
on 3, and the load on 3 will then be greater than 12000 lbs. ; 
for whatever load there is between 3 and 5 is supported at 
3 and 5. 

Moreover, we know that the effect of increasing the load on 
5 is to increase the shearing-force, provided we do not at the 
same time increase that on 3 so much as to destroy the effect 
of increasing that on 5. 

Hence, there must be some point between 3 and 5 to which 
the moving load must extend in order to render the shearing- 
force in panel 3-5 a maximum. 

Let the distance of this point from 5 be x ; then, if we let 

23000 . _ . 

w = — — = moving load per foot of length, 

Moving load on panel = wx, 

Part supported at 3 = , 

20 

TJDX^ 

Part supported at 5 = wx . 

20 

Hence, portion of shearing-force due to the moving load on 
panel 3-5 equals 



CONCENTRATING THE LOAD AT THE JOINTS. 



20; 



12/ 

— ( wx 

i4\ 



wx 2 



\ I wx 2 



20 / 14 20 



w f 1 -zx 2 \ 

i4\ 20 / 



This becomes a maximum when its first differential co-efficient 
becomes zero, i.e., when 

12 — fljs? = o; 

therefore 

* == 9 r .23. 

Hence, when the moving load extends to a distance of 9.23 feet 
from 5, then the shearing-force in panel 3-5, and hence the 
stress in diagonal 3-4, is a maximum. 



Panels. 


Portion of Shearing-Force 

due to Moving Load on 

Panel. 


Value 

of x, in 

feet. 


Portion of Load 

at Joints named 

below. 


Portion of Load 

at Joints named 

below. 


1- 3 
3- 5 
5- 7 
7- 9 
9-1 1 
11-13 
13-15 


27-28 
25-27 
23-25 
21-23 
19-21 
17-19 
15-17 


E( iax - 1^1) 

I4\ 20 / 

™( lIX _ i&) 

14V 20 / 

»( I0X - un) 

14V 20 / 
wf I3^ 2 \ 
14V 20 / 

w (sx - l3 * 2 ) 
14 V 20 / 

^/ lx - I3£f\, 

I4\ 20 / 


IO.OO 

9-23 

8.46 

7.69 
6.92 
6.I5 

5.38 


I 

3 

5 
7 

9 
11 

13 


ii 500 

9797 
8230 
6801 

5507 
4350 
3329 


3 
5 
7 
9 
11 

13 
15 


1 1 500 
1 1432 
11227 
10886 
10409 
9795 1 

9045 ; 



To show how the adoption of this method would affect the 
resulting stresses in the diagonals and verticals, I have given 
the work above, and shown the difference between these and 



206 



APPLIED MECHANICS. 



the former results. In this table x = distance covered by load 
from end of panel nearest the centre. 



Panels. 


Greatest Shearing-Force of Same Kind as that due to Dead Load. 


i- 3 


27-28 


35000 

-^-(i+... + i3) =227500 


3" 5 


25-27 


^p(i +. . .+ 1 1) +^(34932)~^(2i797) = 193385 


5- 7 


23-25 


^p(i+. . .+ io) + ^(34727)-~(202 3 o)-^(i20oo) =161038 


7- 9 


21-23 


3 -ffd+...+ 9)+^(34386)-f 4 (i88oi)-^°(i + 2) =130461 


9-i i 


19-21 


^(1+...+ 8)+^(3 3Q o 9 )-^(i7507)-^f (1 + 2+3) =101654 


11-13 


17-19 


3 -fP(i + ...+ 7)+7 4 (33295)-7 4 (i635o)-^(i+...+4)= 746i6 


13-15 


15-17 


3 -^(i+...+ 6)+f(32545)-7/r5329)+ i ^(i+...+5)= 49345 
14 14 14 14 



Hence, for stresses in main braces, we have 



Diagonals. 


Stresses. 


I- 2 


28-26 


-321685 


3- 4 


27-24 


-273446 


5-6 


25-22 


— 227708 


7-8 


23-20 


— 184472 


9-10 


21-18 


-143739 


11-12 


19-16 


-IO5507 


13-M 


17-14 


— 69774 



Moreover, for the shearing-forces of opposite kind from 



CONCENTRATING THE LOAD AT THE JOINTS. 



207 



those due to dead load, we have, if x = distance from end of 
panel nearest support which is covered by moving load, — 



Panels. 


Portion of Shear due 
to Moving Load on Panel. 


Value 
of X. 


Portion of Load 

at Joints named 

below. 


Portion of Load 

at Joints named 

below. 


13-15 
II-I3 


17-15 
19-17 


™( 6x - 12?) 

14V 20 / 
I4\ 20 / 


4.62 
3-84 


15 
13 


2455 
1695 


13 
II 


8I7I 
7137 



Panels. 



13-15 
11-13 



17-15 
19-17 



Greatest Shearing-Forces of Opposite Kind from those due to Dead Load. 



^(i + ...+5)4-f 4 (3r67i)-f 4 ( I4 455)-~(i+...+6) = 25846 
^(i+...+ 4 )+f 4 (30637)-^(i3695)- £ f > -°(i + .-.+7) = 4"6 



Hence we have the following as the stresses in the counter- 
braces : — 



Counter-Braces. 


Stresses. 


15-12 
I3-IO 


T5-16 

17-18 


- 36546 

- 5820 



And, for the verticals, we have the new, instead of the old, 
shearing-forces. 



zo8 



APPLIED MECHANICS. 



The following table compares the results :- — 



Diagonals. 


Stress, Ordinary 
Method. 


Stress, New Method. 


Difference. 


I- 2 


28-26 


-321685 


-321685 




3- 4 


27-24 


-274518 


-273446 


IO72 


5" 6 


25-22 


-229675 


— 227708 


I967 


7- 8 


23-20 


-187153 


-184472 


268l 


9-10 


21-18 


-I46956 


~!43739 


3217 


1 1-12 


19-16 


— IO9080 


- io 55°7 


3573 


13-H 


17-14 


- 735 2 8 


- 69774 


3754 


15-12 


15-16 


— 4O299 


-3 6 546 


3753 


13-10 


17-18 


- 9393 


- 5820 


3573 



Verticals. 


Stress, Ordinary 
Method. 


Stress, New Method. 


Difference. 


3- 2 


27-26 


+ 2275OO 


4-227500 


O 


5- 4 


25-24 


+ I94M3 


+ 193385 


758 


7- 6 


23-22 


+ 162429 


+ 161038 


1391 


9-8 


21-20 


+ J 3 2 357 


+ 130461 


1896 


II-IO 


19-18 


+ 103929 


+ IO1654 


2275 


! 13- 12 


17-16 


+ 77143 


+ 74616 


2527 


15-14 


4- 28500 


+ 49345 ' 


2655 



§156. Compound Bridge-Trusses. — The trusses already 
discussed have contained but a single system of latticing, or 



COMPOUND BRIDGE-TRUSSES. 



209 



at least only one system that comes in play at one time ; so that 
a vertical section never cuts more than three bars that are in 
action simultaneously, the main brace having no stress upon it 
when the counterbrace is in action, and vice versa. 

We may, however, have bridge-trusses with more than one 
system of lattices ; and, in determining the stresses in their 
members, we must resolve them into their component systems, 
and determine the greatest stress in each system separately, 
and then, for bars which are common to the two systems, add 
together the stresses brought about by each. 

In some cases, the design is such that it is possible to 
resolve the truss into systems in more than one way, and then 
there arises an uncertainty as to which course the stresses will 
actually pursue. 

In such cases, the only safe way is to determine the greatest 
stress in each piece with every possible mode of resolution of 
the systems, and then to design each piece in such a way as to 
be able to resist that stress. 

Generally, however, such ambiguity is an indication of a 
waste of material ; as it is most economical to put in the bridge 
only those pieces that are absolutely necessary to bear the 
stresses, as other pieces only add so much weight to the struc- 
ture, and are useless to bear the load. 

The mode of proceeding can be best explained by some 
examples. 

Example I. — Given the lattice-girder shown in Fig. 117, 
loaded at the lower panel points 



2 4 6 8 10 12 14 16 18 20 22 34 2%j^ 



1 3 5 7 9 11 13 15 17 19 21 23 25 

only. Dead load = 7200 lbs. 

per panel point, moving load 

= 18000 lbs per panel point; 

let the entire length of bridge 

be 60 feet ; let the angle made by braces with horizontal 

= 6o°. 



Fig. 117. 



2IO 



APPLIED MECHANICS. 



This truss evidently consists of the two single trusses shown 
in Figs, wja 



>3 



^ 



«K 



20 24 



11 



15 



23 



10 14 18 
Fig. 117J. 



the 



and nyb ; and 
we can compute 
the greatest 
stress of each FlG ' " 7 "' 

kind in each member of these trusses, and thus 
obtain at once 
all the diag- 
onal stresses, 
and then, by 
addition, the 
greatest chord stresses. 

Thus the stress in 1-3 (Fig. 117) is 
same as the stress in 1-5 (Fig. 117^). 

The stress in 3-5 = stress in 1-5 (Fig. 
117^) + stress in 3-7 (Fig. 117^). 

The stress in 5-7 = stress in 5-9 (Fig. 
117*2) + stress in 3-7 (Fig. Wjb). 

The results are given on the diagram (Fig. 
117c); the work being left for the student, as 
it is similar to that done heretofore. 

Example II. — Given , the lattice-girder 
shown in Fig. 118. Given, as before, Dead 
load = 7200 lbs. per panel point, moving load 
= 18000 lbs. per panel point, entire length of 
bridge = 25 feet ; load applied at lower panel 
points. 



Solution. — In this case, there are two possible modes of 
resolving it into systems. The first is shown in Figs. 118a and 
\\2>b: and this is necessarily the mode of division that must 
hold whenever the load is unevenly distributed, or when the 



COMPO UND BRID GE- TR USSES. 



211 



travelling-load covers only a part of the bridge ; for a single 
load at 6 is necessarily put in communication with the support 
at 2 by means of the diagonals 6-3 and 3-2, and with the sup- 
port at 12 by means of the diagonals 6-y, 7-10, io-n, and the 
vertical 11-12, and can cause no stress in the other diagonals 



5 7 9 11 



2 4 6 8 10 12 
Fig. 118. 



2 6 10 12 

Fig. 118a. 



2 4 8 12 

Fig. ii8<5. 



* 5 7 n 

24 10 12 

Fig. tiSc. 



5 7 9 



AXA 

2 6 8 12 

Fig. u8d. 



When, however, the whole travelling-load is on the bridge, 
it is perfectly possible to divide it into the two trusses shown 
in Figs. 118c and 118^, the diagonals 4-5, 7-10, 6-y, and 5—8 
having no stress upon them. 

When the load is unevenly distributed, we have certainly 
the first method of division ; and when evenly, we are not sure 
which will hold. 

Hence we must compute the greatest stresses with each 
mode of division, and use for each member the greatest ; for 
thus only shall we be sure that the truss is made strong 
enough. 

We shall thus have the following results : — 



212 



APPLIED MECHANICS. 



FIRST MODE OF DIVISION (FIGS. 1x80 AND n8£). 



Diagonals. 


Greatest Shearing-Force 
oi One Kind. 


Greatest Shearing-Force 
of Opposite Kind. 






Fig. 
1 \Za. 


Fig. 
n8£. 


Corresponding 
Stresses. 


2- 3 


12-9 


^~(3 + l ) — 20l6o 


O 




+23279 


— O 


3-6 


9-3 


^(3 + 1) = 20160 


O 




-23279 


+ O 


6- 7 


8-5 


2 J^_7Joo (2)= 2j6o 
5 5 • ' 


25200/ v _ 7£^2 _ 
5 W 5 


864O 


+ 2494 


1 
- 9976 


7-10 


5-4 


5 5->_7Joo 2j6o 

5 5 V ' 


25200, . 7200 
5 ' 5 _ 


864O 


— 2494 


+ 9976 


j IO-II 


4-i 


O = 


^(3 + 4) = 




30240 


O 


-34918 



Chords. 
Supporting force at 2 (Fig. 118V) or 12 (Fig. n8£) 

= 2J f(3 + 1) = 20160, 
Supporting force at 12 (Fig. n 8a) or 2 (Fig. u8£) 

= '-^{2 + 4) = 30240. 



j Section. 

1 


Bending-Moment. 


Chords. 


Maxi- 
mum 

Stresses 
in 

Separate 






Com- 
ponents 

of 
Stresses. 


Greatest 

Resultant 

Stresses. 


1 M 


•0" 
00 


« 


00 


Choras. 


i to 


9 




ha 
E 


bi j Trusses. 






j 3 


201 60 X 5 = 100800 


2- 6 


8-12 


-IT639 


1-3 


j 

9-1 ij 0+I-5 


+ 17459 


1 6 


8 


20160 X 10=201600 


3- 7 


5- 9 


+23279 


3-5 


7- 9.3- 7+1-5 


+40738 


1 7 


5 


20160X15 — 25200 










1 
| ■ 




| 




x 5=176400 


6-10 


4- 81—20369 


5-7 


13- 7+5-9 


+46558 


! IO 


4 


30240X 5 = 151200 


7-i 1 


1- 5I+17459 


2-4 


IO-I2 ( 2- 6-J-2-4 


-II639I 


! 






10-12 


2- 4 

i 


4-6 
6-8 


8-IQ2- 6-I-4-8 
6-10+4-8 


— 32008 | 

-40738 ! 

I 



COMPOUND BRIDGE-TRUSSES. 



213 



SECOND METHOD OF DIVISION (FIGS. 118* AND n&f). 
Diagonals (Fig. n8<r). 



Diagonals. 


Maximum 
Shear. 


Corresponding 
Stresses. 


1-4 

4-5 


IO-II 

7-10 


252OO 
O 


— 29O98 
O 



Fig. 11 8a 7 . 



Diagonals. 


Maximum 
Shear. 


Corresponding 
Stresses. 


2-3 
3-6 
6-7 


9-12 

8- 9 
5-8 


252OO 

252OO 

O 


+ 29O98 

— 29098 

O 



Chords. 

Each supporting force in either figure = 25200. 
Fig. n8<:. 

Bending-moment anywhere between 4 and 10 = (25200) (5) = 126000; 

.*. Stress in 1-11 = -f- 14549, 
.-. Stress in 4-10 = —14549. 

Fig. 118^. 

Bending-moment at 3 or 9 = 126000, 

Bending-moment anywhere between 6 and 8 = 252000; 

.*. Stress in 3-9 = +29098, 

Stress in 2-6 or 8-12 = —14549, 
Stress in 6-8 = —29098. 



214 



APPLIED MECHANICS. 



Hence we have for chord stresses, with this second divis- 
ion, — 



Chords. 




Stresses. 


i-3 


9-1 1 


I-II.+ o 


+ 14549 


3-5 


7- 9 


i-n + 3-9 


+ 43647 


5-7 


- 


i-ii + 3-9 


+ 43647 


2-4 


IO-I2 


O -f- 2-6 


-14549 


4-6 


8-io 


4-10 4- 2-6 


— 29098 


6-8 


— 


4-10 4- 6-8 


-43647 



Hence, selecting for each bar the greatest, we shall have, as 
the stresses which the truss must be able to resist, — 



1-4 


10-11 


+ 


-34918 


i-3 


9-1 1 


+ 17459 


2-3 


12-9 


4-29098 


— 


3-5 


7" 9 


+43647 


3-6 


9 -8 


+ 


— 29098 


5-7 


- 


+46558 


4-5 


10- 7 


+ 9976 


— 2494 


2-4 


10-12 


-14549 


5-8 


7- 6 


+ 2494 


- 9976 


4-6 
6-8 


8-10 


— 32008 
—43-647 



These results are recorded in Fig. n8e. 



(l)+17459(8)+43647(5)-M6558(?)+43647(9W459ftfl 




(2*-U549<4)- 32008 (6) - 43647(.8)-32008(10)-14549a2) 
Fig. 118*. 



§157. Other Trusses. — In Figs. 119, 120, and 121, we 
have examples of the double-panel system with the load placed 



OTHER TRUSSES. 



at the lower panel points only. When, as in 119 and 120, the 
number of panels is odd, the same ambiguity arises as took place 
in Fig. 118. When, on the other hand, the number of panels 
is even, as shown in Fig. 121, there is only one mode of division 
into systems possible. The diagrams speak for themselves, and 
need no explanation. 




11 13 15 IT 19 21 
Fig. 119. 



25 27 29 31 33 34 





31 33 34 



2 4 




22 



26 



30 32 




17 19 
Fig. 119c. 



27 31 33 34 




24 28 




7 11 15 21 35 29 S3 
Fig. iigrf. 



2l6 



APPLIED MECHANICS. 




2 4 t 


12 16 20 


24 28 32 36 


\ 




\ 






\ 

\ 








/ 


\ 




\ 




/ \ 




\ 






/ 



13 7 11 



'19 
Fig. 120a. 



27 



6 10 14 18 22 26 30 34 






15 9 



13 17 21 

Fig. i2o3. 



25 29 33 35 



2 6 10 14 



24 28 32 36 



1 5 



17 19 
Fig. i2oc. 



23 27 31 35 



2 4 8 12 



26 30 34 36 



1 3 



11 15 21 25 

Fig. wod. 



2 4 6 8 10 12 14 16 18 






1 3 5 7 9 11 13 15 17 19 20 
Fig. 121. 



FINK'S TRUSS. 



21/ 




2 4 


3 12 16 18 


/ 








A 


1 


i 


» 13 1 


7 20 



Fig. i2i<5. 



Fig. 122. 



The trusses given above may be considered as examples, to 
be solved by the student by assuming the dead and the moving 
load per panel point respectively. 

§158. Fink's Truss. — The description of this truss will 
be evident from the figure. There is, first, the primary truss 
1-8-16; then on each side 
of 9-8 (the middle post of 
this truss) is a secondary 
truss (1-4-9 on th e left* 
and 9-1 2-16 on the right). 

Each of these secondary 
trusses contains a pair of smaller secondary trusses, and the 
division might be continued if the segments into which the 
upper chord is thus divided were too long. 

Of the inclined ties, there is none in which any load tends 
to produce compression ; in other words, every load either in- 
creases the tension in the tie, or else does not affect it. Hence 




Fig. 123. 



218 



APPLIED MECHANICS. 



the greatest stresses in all the members will be attained when 
the entire travelling-load is on the truss, and we need only con- 
sider that case. 

The determination of the stress in any one member can 
readily be obtained by determining, by means of the triangle 
of forces, the stress in that member due to the presence of 
the total load per panel point, at each point, and then adding the 
results. This will be illustrated by a few diagonals. 

Let angle 8-1-9 — 2 > 
Let angle 4-1-5 = i" x , 
Let angle 2-1-3 — h > 

we shall have, if w + w I = entire load per panel point, — 



c 
.2 • 






Effect of Loads at 






in 

c 


Q 














5 '3 


3 


5 


7 


9 


11 


13 


15 


1-2 


W + Wi 

2 sin i 2 




















^ + ze/i 
2 sin z 2 


2 -5 


W + Wi 

2 sin i 2 




















«/ + Wj 


2 sin z 2 


5-6 








W + W-i 

2 sm z 2 














W + Wi 

2 sin z 2 


6-9 








ze/ + w-i 
2 sin z 2 














2 sin z 2 


i-4 


W + Wi 

4 sin i x 


W + Wi 

2 sin i t 


w -h w I 
4 sin z x 














w 4- 7^i 


sin /i 


4-9 


W + Wi 

4 sin ij 


7V + Wi 

2 sin i 1 


7# + Wi 

4 sin z'i 














w -\- W-i 
sm z t 


1-8 


w + a/, 
8 sin i 


w -\- W-i 
4 sin i 


3 7f-}- W » 


W + Wi 

2 sin 1 


3 w-\-w x 


W 4" Wy 

4 sin i 


w 4- Wi 
8 sin i 


2(w-f-7e/i ) J 


8 sin 2 


8 sin i 


sin 1 1 



The stresses in all the other members may be found in a 
similar manner. 



GENERAL REMARKS. 



219 



§ 159. Bollman's Truss. — The description of this truss is 
made sufficiently clear by the figure. The upper chord is made 
in separate pieces ; and 
the short diagonals 2-5, 
3-4, 4-7, 5-6, 7-8, 6-9, 
8-1 1, and 9-10 are only 
needed to prevent a 
bending of the upper 
chord at the joints. 
When this is their only object, the stress upon them cannot be 
calculated : indeed, it is zero until bending takes place ; and 
then it is the less, the less the bending. Hence, in this case, 
the stress is wholly taken up by the principal ties ; and these 
have their greatest stress when the whole load is on the bridge. 

The computation of the stresses is made in a similar man- 
ner to that used in the Fink. 




§160. General Remarks. — The methods already explained 
are intended to enable the student to solve any case of a bridge- 
truss where there is no ambiguity as to the course pursued by 
the stresses. 

In cases where a large number of trusses of one given type 
are to be computed, it would, as a rule, be a saving of labor to 
determine formulae for the stresses in the members, and then 
substitute in these formulae. 

Such formulae may be deduced by using letters to denote 
the load and dimensions, instead of inserting directly their 
numerical values ; and then, having deduced the formulae for 
the type of truss, we can apply it to any case by merely sub- 
stituting for the letters their numerical values corresponding 
to that case. 

Such sets of formulae would apply merely to specific styles 
of trusses, and any variation in these styles would require the 
formulae to be changed. 



220 APPLIED MECHANICS. 

In order to show how such formulae are deduced, a few will 
be deduced for such a bridge as is shown in Fig. in. 

Let the load be applied at the upper panel points only ; let 
dead load per panel point = w, moving load per panel point 
= w z . Let the whole number of panels be N f N being an even 
number. Let the length of one panel = height of truss == /. 
Then length of entire span = Nl. 

Consider the {n -f- i) Ul panel from the middle. 

The stress in the main tie is greatest when the moving load 
is on all the panel points from the farther abutment up to the 
panel in question, {n -\- i) th . 

Hence, for the n th panel from the middle, the greatest shear' 
ing-force that causes tension in the main tie is equal to 

w~\-w 



..{ I+2+3+ ... + (f + «)[-^ I+2+ 3 + ... + (f-«- I )[ 

= 2^V"1 Wl \T~*~ 7 J r— J r n \+ wi < 2n + 1 ) N \ 
Hence stress in main tie 

w\ ( \- n\ -\ 1- n + w{2n -f- i)iV^ !■ . (i) 



2N 

For the counterbrace, we should obtain, in a similar way, the 
formula 



V7 



2JV 



'N 


V 


N i 1 




n) 


+ n 


K 2 


J 


2 



— wN(2n + i) > 



which represents tension when it is positive. Proceed in a 
similar way for the other members. 

When there is more than one system, we must divide the 
truss into its component systems; and when there is ambiguity, 
we must use, in determining the dimensions of each member, 
the greatest stress that can possibly come upon it. 



CENTRE OF GRAVITY. 221 



CHAPTER V. 

CENTRE OF GRAVITY. 

§ 1 6 1. The centre of gravity of a body or system of bodies, is 
that point through which the resultant of the system of parallel 
forces that constitutes the weight of the body or system of 
bodies always passes, whatever be the position in which the 
body is placed with reference to the direction of the forces. 

§ 162. Centre of Gravity of a System of Bodies. — If 
we have a system of bodies whose weights are W„ W 2 , W v etc., 
the co-ordinates of their individual centres of gravity being 
(*„ y» #,), (^2, y» z 2 ), (* v y v # 3 ), etc., respectively, and if we 
denote by x oy y , z oy the co-ordinates of the centre of gravity of 
the system, we should obtain, just as in the determination of the 
centre of any system of parallel forces, — 

i°. By turning all the forces parallel to OZ, and taking 
moments about OY, 

{W, + W 2 -f- W 3 + etc.)* = W,x L + W 2 x 2 + W % x z + etc., 

or 

• x ^W = %Wx; 

and, taking moments about OX, 

(W x + W 2 + IV 3 + etc.)y Q = W x y x + W 2 y 2 + W 3 y 3 + etc., 

or 

y Q %W = ^Wy. 



222 



APPLIED MECHANICS. 



2°. By turning all the forces parallel to OX, and taking 
moments about OY, 



or 



(W, 4- W % + W z + etc.)*, = W lZl + ^ + 0^ + etc., 



s 2J^ = W*. 



Hence we have, for the co-ordinates of the centre of gravity 

of the system, 

%Wx %Wy %Wz 

y Q — -^ttv, z n = 



2,1V' 



2JV' 



%W 



EXAMPLES. 

I. Suppose a rectangular, homogeneous plate of brass (Fig. 125), 

where AD =12 inches, AB = 5 inches, 
and whose weight is 2 lbs., to have 
weights attached at the points A, B, C, 
and D respectively, equal to 8, 6, 5, and 
3 lbs. ; find the centre of gravity of the 
system. 

Solution, 







Y 




D 






A 




O 


f 












C 






B 



Fig. 125. 



Assume the origin of co-ordinates at 



the centre of the rectangle, and we have 



W z =2, W 2 = 8, W 3 = 6, W 4 = 5, W s = 


3. 


v-V j ■ ' ^-'5 l ^*'2 ' V*« «A"» - ■ L/» «^4 — \J » «^v r ■■■ — 


-6, 


J>, =0, ^ 2 = f, ^3 = -f, ^ 4 = -f, J 5 = 


f.i 


.\ S^c = -f 48 + 36 — 30.0 — 18.0 = 36, 




%Wy = + 20 — 15 — 12.5 + 7.5 = 0, 




2W =2+ 8-f 6-f- 5.0+ 3.0= 24; 




36 





Hence the centre of gravity is situated at a point E on the line OX, 
where OE =1.5. 




2. Given a uniform circular plate of radius 8, and weight 3 lbs. 
(Fig.. 126). At the points A, B, C, and D, 

weights are attached equal to 10, 15, 25, and 23 o ■■ ' 
lbs. respectively, also given AB = 45 °, BC = [ n ^a 

105 °, CZ> = 120 ; find the centre of gravity of 
the system. 



§ 163. Centre of Gravity of Homogeneous Bodies. — For 
the case of a single homogeneous body, the formulae have been 
already deduced in § 44. They are 

__ fxdV __ fydV fzdV 

x °-~Sdv> ^"Tdv 1 z °~Jdv' 

and for the weight of the body, 

W = wfdV, 

where x , y oi z oy are the co-ordinates of the centre of gravity of 
the body, W its weight, and w its weight per unit of volume. 

From these formulae we can readily deduce those for any 
special cases ; thus, — 

(a) For a volume referred to rectangular co-ordinate axes, 
d V = dxdydz. 

__ fffxdxdydz __ fffydxdydz __ fffzdxdydz 

fffdxdydz' ° fffdxdydz' ° fffdxdydz' 

(b) For a flat plate of uniform thickness, t, the centre of grav- 
ity is in the middle layer ; hence only two co-ordinates are 
required to determine it. If it be referred to a system of rect« 
angular axes in the middle plane, dV '= tdxdy, 

_ ffxdxdy _ ffydxdy 

** f/dxdy' y ° SSdxdy 



224 APPLIED MECHANICS. 

The centre of gravity of such a thin plate is also called the 
centre of gravity of the plane area that constitutes the middle 
plane section ; hence — 

(c) For a plane area referred to rectangular co-ordinate axes 
in its own plane, 

= ffxdxdy _ ffydxdy 

° ffdxdy' y ° ffdxdy' 

{d) For a slender rod of uniform sectional area, a, if x, y, 2, 
be the co-ordinates of points on the axis (straight or curved) of 
the rod, we shall have dV — ads — a\l(dx) 2 -+- (dy) 2 + (dz) 2 t 



fxds 



v _ Syds 
y °~7dl 



fzds 

Zo = -7-7- 

fds 









(e) For a slender rod whose axis lies wholly in one plane, 

the centre of gravity lies, of course, in the same plane ; and if 

our co-ordinate axes be taken in this plane, we shall have 2 = G 

d2 

— = o, and also 2 Q = o. Hence we need only two co- 

ax 



CENTRE OF GRAVITY OF HOMOGENEOUS BODIES. 225 

ordinates to determine the centre of gravity, hence dV '= ads 
= a\/(dx)' + (dy)\ 









Ids 



/v / - + (D- 



(/) For a line, straight or curved, which lies entirely in one 
plane, we shall have, again, 



_ fxds 

Xo ~7dI 






Whenever the body of which we wish to determine the 
centre of gravity is made up of simple figures, of which we 
already know the positions of the centres of gravity, the method 
explained in § 162 should be used, and not the formulae that 
involve integration ; i.e., taking moments about any given line 
will give us the perpendicular distance of the centre of gravity 
from that line. 

In the case of the determination of the strength and stiff- 
ness of beams, it is necessary to know the distance of a hori- 
zontal line passing through the centre of gravity of the section, 



226 APPLIED MECHANICS. 

from the top or the bottom of the section ; but it is of no prac- 
tical importance to know the position of the centre of gravity 
on this line. In most of the examples that follow, therefore, 
the results given are these distances. These examples should 
be worked out by the student. 

In the case of wrought-iron beams of various sections, on 
account of the thinness of the iron, a sufficiently close approxi- 
mation is often obtained by considering the cross-section as 
composed of its central lines; the area of any given portion 
being found by multiplying the thickness of the iron by the 
corresponding length of line, the several areas being assumed 
to be concentrated in single lines. 

EXAMPLES. 

i. Straight Line AB (Fig. 127). — The centre of gravity is evidently 
at the middle of the line, as this is a point of 
A 1 - B symmetry. 

Fig. 127. 

2. Combination of Two Straight Lines. — The centre of gravity in 
each case lies on the line 00^ Figs. 128, 129, 130, and 131. 

(a) Angle-Iron of Unequal Arms (Fig. 128). — Length AB = b, 
length BC — h, area AB = A, area 

BC = B; 



bh _ 

.-. BE = DE = i , 

(b) Angle-Iron of Equal Arms (Fig. 129). — Length AB = BC 

B = b; 

... BE = DE = ^ n ^ 2 . 

Fig. 129. 





CENTRE OF GRAVITY OF HOMOGENEOUS BODIES. 22/ 



(c) Cross of Equal Arms (Fig. 130). — AB *= 00 s = h; 



h 

AC= BC = - 

2 



B 

Fig 130. 

(d) T-Iron (Fig. 131). — Area AB = A, area CE = B, length 
A e b CE =. h; 

Bh 



DE = 



c 

Fig. 131. 



2{A + B)' 



3. Combination of Three Lines, — 00 t = line passing through the 
centre of gravity. 

(a) Thin Isosceles Triangular Cell (Fig. 132). — Length AB = 
^C = a, length AC = b, area ^# = ^C a d c 

= A, area AC — B ; 

.'. DB 



-1*-\ 



= JV(2« - £)(2* + J) 
A 




DE = 



2(2^ + ^) 
^ -f ^ 



2(2^ -f B) 
(b) Same in Different Position (Fig. 133) 



Sj(2a - b)(2a + b), 



Sj(2a -b)(2a + t). 




BD = DC = - 



Fig. 113. 



228 



APPLIED MECHANICS. 



(c) Channel-Iron (Fig. 134). — Area of flanges = A, area of web 

= B, depth of flanges -f- \ thickness of 

A „ web = h - 



• cp-1 Ah 



Fig. 134. 



DE = - 



2 A + B 

(d) 1-Beam (Fig. 135). — Area of upper flange = A 19 area of 
lower flange = A 2 , area of web = B, height = h. 

h 2A 2 + B 

— — ? 

o— 



CG = 



2 A 1 + A 2 + B 



„^ h 2 A, + B 

GD = - 



C B 
G 



O, 



2 A z + A 2 + B 



E D F 

Fig. 135. 



4. Combination of Four Lines. — 00 I = line passing through the 
centre of gravity. 

(a) Thin Rectangular Cell (Fig. 136). — Length AB = h; 

h 



D A 
F E 



C B 



-o, 



AE = BE = 



Fig. 136. 

(b) Thin Square Cell (Fig. i^). — AB = BC = h; 

h 



BE = CE = 



A B 

F E 



-0. 



Fig. 137. 



5. Circular Arcs. 

(a) Circular Arc AB (Fig. 138). — Angle A OB = 6 If radius = r, 



Use formula 




b 



x c 



Jxds 
~Jds> 



fyds 

Ids' 



but use polar co-ordinates, where 
Fig. i 3 s. ds = rdO, x = rcosO, y — rsintf, 



CENTRE OF GRAVITY OF HOMOGENEOUS BODIES. 229 



X = 



■£■ 



_ -(sing,) 



r* J cosy 

r 2 / sin&/0 , .. . , 1Z1 

Jo (1 - cos0,) sin 2 J0, 

y Q - — - = ; ^ 2 r- 






(b) Circular Arc AC (same figure). 



x a = 



r sin $! 



1 — > Jo = o. 
(c) Quarter-Arc of Circle AB, Radius r (Fig. 139). 



x = 



■1 



>- 2 I cos 



■i 



2r 

IT 



y Q 



\ dO 
2r 



0* 



FlG.iaf. 



(d) Semi-circumference ABC (same figure), 



x = —, y Q = o. 

6. Combination of Circles and Straight Lines. 

Barlow Rail (Fig. 140). — Two quadrants, radius r, and web, 
c , whose area = -£■ the united area of the quadrants. 
Let united area of quadrants = A, area of web 




= -&-A; let EF = x 



... HAxo = A(£) = ftAt 



= ¥, 



*o = - = -£/?,. 



230 



APPLIED MECHANICS. 



7. Areas. 

(a) T-Section (Fig. 141). — Let length AB = B, EF = b, entire 
height = H, GE = h. Let distance of centre 
of gravity below AB = # x ; therefore, taking 




1 moments about AB as an axis, 
x x \BH - h{B - b)} 

= \BH* - h{B - b)\H - k \ 



whence we can readily derive x s 



(b) I-Section (Fig. 142). — Let AB = B, GH = b, MN = b l9 
entire height = U, BC = If — h, EH = A, ; and let # r = distance 
of centre of gravity below AB. A b 

Hence, taking moments about AB, we have 

x x \B{H - h) + b x {h - k z ) + bh x \ 

= *{H- hy+bhlH-fi 

I h - h x 

+ ^ _ W H - h + — j— * 

whence we can deduce #,. 




(r) Triangle (Fig. 143). — If we consider the triangle OB C as 
composed of an indefinite number of narrow strips parallel to the side 
CB, of which FLHK is one, the centre of gravity 
of each one of these strips will be on the line OD 
drawn from O to the middle point of the side 
CB ; hence the centre ^f gravity of the entire tri- 
angle must be on the line OD.' For a similar rea- 
son, it must be on the median line CE ; hence the 
centre of gravity must be at the intersection of the median lines, and 
hence 

BC. ODsinODC 
x Q = OG = fOD. Moreover, area = > 





CENTRE OF GRAVITY OF HOMOGENEOUS BODIES., 23 1 
(d) Trapezoid (Fig. 144). 

First Solution. — Bisect AB in O, and CE in Dj let g x be the 

centre of gravity of CEB, and g 3 that of ABC. 
Then will G, the centre of gravity of the trape- 
zoid, be on the line g^^ and 

G gl ABC 
Gg 2 CEB' 

But it must be on the line ODj hence it is at their intersection. 
From the similarity of GG 1 g l GG 2 g^ we have 



GG X G gl ABC A^B = B # 
GG 2 ~ Gg, ~ BEC CE b ' 



GG 2 , . nn OD 

g£ = b+1' andsmce ^ 2 = T' 



-. OG=OG.+ GG t = ^+GG. = ^[i +:ff ^j. 



Second Solution. — Fig. 144 (a). Let be the 
point of intersection of the non-parallel sides AC 
and BE. Let OF = a?„ OP = # 2 , OG = x . Take 
moments about an axis through 0, and perpen- 
dicular to OF, and we readily obtain 



3 s 

_ 2 X i — X a 
X — jj ^. 

3 ■*! — *, 




232 



APPLIED MECHANICS. 



(e) Parabolic Half- Segment OAB (Fig. 145). — Let OA = x S9 
AB — yi ; let x 0l yo-, be the co-ordinates of the centre of gravity ; let 
the equation of the parabola be f = \ax : 



/ / xdxdy 2a* I x*dx 



■»onJ -r^ 



J dxdy 2 a? I x^dx 



5 

T X J_ _ 3 



Fig. 145. 



j-*i, 



Jo I y dxd y , k _ 



(/) Parabolic Spandril OBC (Fig. 145). — Let #0, J^o, be co-ordi- 
nates of centre of gravity of the spandril. 



/ / , xdxdy 
I I dxdy 

y jdxdy 



— TU x i> 



y = 



fy» 



Area = xj x — \xj, — \x,y v 



CENTRE OE GRA VI TY OF HOMOGENEOUS BODIES. 233 



(g) Circular Sector OAC (Fig. 146).— Let OA — r, AOX — 0„ 
•*o, ^o, be the co-ordinates of the centre of gravity : 



■ Jc 



'♦V*-jr» 



nr p* r^-xi /Vcosfl, f*x tan 0, 

/ / xdxdy -\- I I xdxdy 



0, ' 



Area = \r(2r6 x ) = r 2 6,. 



c 

Fig. 146. 



Second Solution. 

Consider the sector to be made up of an indefinite number of 
narrow rings ; let p be the variable radius, and dp the thickness : 

.'. Elementary area = 2pd 1 dp, 

and centre of gravity of this elementary area is on OX, at a distance 

from equal to p — ^— ■ [see Example 5 (^)] ; 



£ I "nr 1 \ ^ 2p ^ dp * 2 sin '■ jf^* 

(A) Circular Half -Segment ABX (Fig. 146), 



sin 



=«*v 



*0 = 



/ xdxdy I x Vr* — jcVjc 

r COS*,* 7 ° _ * r COS <>! 



Sector minus triangle \r % B x — -|V sin X cos 0, 



in 8 0. 



sin 



= *r 



0, — sin 0, cos 6* 



Vr* - X* 






r cos B x 



y ° ~ £^(0, - sin 0, cos 0,) _ * 0, - sin 0, cos 



= k : 



4 sin'^fl, — sin'fl, cos 0, 



234 APPLIED MECHANICS. 

§ 164. Pappus's Theorems. — The following two theorems 
serve often to simplify the determination of the centres of 
gravity of lines and areas. They are as follows : — 

Theorem I. — If a plane curve lies wholly on one side of a 
straight line in its own plane, and, revolving about that line, 
generates thereby a surface of revolution, the area of the sur- 
face is equal to the product of the length of the revolving line, 
and of the path described by its centre of gravity. 

Proof. — Let the curve lie in the xy plane, and let the axis 

of y be the line about which it revolves. We have, from what 

fxds 
precedes, § 163 (e), x = -jjp 

.". x fds = fxds, 

where x equals the perpendicular distance of the centre of 
gravity of the curve from OY, ds = elementary arc, 

2-rrx fds = f(2irx)ds; 

or, reversing the equation, 

f{2lTX)ds = (2TTX )S. 

But f(2irx)ds = surface described in one revolution, while s = 
length of arc, and 2wx = path described by the centre of 
gravity in one revolution. Hence follows the proposition. 

Theorem II. — If a plane area lying wholly on the same 
side of a straight line in its own plane revolves about that line, 
and thereby generates a solid of revolution, the volume of the 
solid thus generated is equal to the product of the revolving 
area, and of the path described by the centre of gravity of the 
plane area during the revolution. 



PAPPUS'S THEOREMS. 235 

Proof. — Let the area lie in the xy plane, and let the axis 
OY be the axis of revolution. We then have, from what has 
preceded, if x Q = perpendicular distance of the centre of gravity 
of the plane area from OY, the equation, § 163 (&), 

ffxdxdy 



x Q = 



Hence 



ffdxdy 



or 



Xoffdady — ffxdxdy; 

•\ (2ttx ) ffdxdy = ff(2TTx)dxdfy 

ff{2irx)dxdy = 2-rrx ffdxdy. 

But ff(2irx)dxdy = volume described in one revolution, and 
2ttx = path described by the centre of gravity in one revolu- 
tion. Hence follows the proposition. 

The same propositions hold true for any part of a revolution, 
as well as for an entire revolution, since we might have multi- 
plied through by the circular measure 0, instead of by 2?r. 

It is evident that the first of these two theorems may be 
used to determine the centre of gravity of a line, when the 
length of the line, and the surface described by revolving it 
about the axis, are known ; and so also that the second theorem 
may be used to determine the centre of gravity of a plane area 
whenever the area is known, and also the volume described by 
revolving it around the axis. 

EXAMPLES. 

1. Circular Arc AC (Fig. 138). — Length of arc = s = 2r6, sur- 
face of zone described by revolving it about OY = circumference of a 
great circle multiplied by the altitude = (27rr)(2rsmO I ); 

.'. (27rx )(2r6 l ) ={2Trr)(2rsin0 1 ) .*. x o 1 = rsin$ l 

smO x 
... x=r- I -. 



236 APPLIED MECHANICS. 

2. Semicircular Arc (Fig. 139). — Length of arc = tit, surface of 
sphere described = 471-r 2 ; 

/. 27rx {7rr) = 471-r 2 .\ jtr = — 

7T 

3. Trapezoid (Fig. 147). — Let ^Z> = b, BC = b; let it revolve 
around ^4Z> : it generates two cones and a cylinder. 

-, AD + BC „ 
Area of trapezoid = BG, 






7r(<£#) 2 
Volume = ~ -{AG + HD) + tt(GB) 2 . BC 



ir(GB)\ 

c = (AG + HD+zBC) 
Fig - *«■ = — (^Z> + ^C + BC) 

o 

... (2 ^) (^±^) . G* = ^j {AD + BC )+ BC\ 
GBf BC \ GBI b \ 

•'■ *° = -tV+ab + W = TV 1 + iT+W = *£ 

4. Circular Sector ACO (Fig. 146). — Area of sector = r*6 If 

volume described = -Jr(surface of zone) = Jr(27rr) (2r sin X ) — 

j^-Trr* sin X ; 

sin X 
.*. (27rx )(r 2 6 I ) = f^ssin^ /. x Q == |r— ^ — 

§ 165. Centre of Gravity of Solid Bodies. — The general 
formulae furnish, in most cases, a very complicated solution, and 
hence we generally have recourse to some simpler method. A 
few examples will be given in this and the next section. 



CENTRE OF GRAVITY OF SYMMETRICAL BODIES. 



'37 



Tetrahedron ABCD (Fig. 148). — The plane ABE, containing the 
edge AB and the middle point E of the edge CD, bisects all lines 
drawn parallel to CD, and terminating in the faces 
ABD and ABC : hence a similar reasoning to that 
used in the case of the triangle will show that the cen- 
tre of gravity of the pyramid must be in the plane 
ABE ; in the same way it may be shown that it must 
lie in the plane ACE. Hence it must lie in their 
intersection, or in the line AG joining the vertex A 
with the centre of gravity (intersection of the medians) 
of the opposite face. In the same way it can be shown that the centre 
of gravity of the triangular pyramid must lie in the line drawn from 
the vertex B to the centre of gravity of the face A CD. Hence the 
centre of gravity of the tetrahedron will be found on the line A G at 
a distance from G equal to \A G. 




Fig. 148. 



§ 166. Centre of Gravity of Bodies which are Symmet- 
rical with Respect to an Axis. — Such solids may be gener- 
ated by the motion of a plane figure, as ABCD 
(Fig. 1*49), of variable dimensions, and of any 
form whose centre G remains upon the axis 
OX ; its plane being always perpendicular to 
OX, and its variable area X being a function 
of x, its distance from the origin. 
/y Here the centre of gravity will evidently 

fig. 149. ij e on t h e ax j s oj{ } anc j t k e elementary vol- 

ume will be the volume of a thin plate whose area is X and 
thickness A;tr; hence the elementary volume will be XAx. 




Take moments about OY, and we shall have 



or 



x Q /Xdx = fXxdx and Volume = fXdx, 

F= fXdx. 



fXxdx 

X = —rr^r-, — i 



fXdx 



2 3 8 



APPLIED MECHANICS. 



EXAMPLES. 

x 2 y 2 z 2 
I. Ellipsoid — -f r- -+- — = i (Fig. 150). — Find centre of gravity 
a D c 

2 of the half to the right of the x plane. Let OK 

= x. Now if, in the equation of the ellipsoid, 

we make y = o, we have — H = 1 : 

a 2 c 2 




Fig. 150. 

where z = EK. 



: z = -Vtf 2 ^-* 2 , 



,# 2 y* 
Make z = o in the equation of the ellipsoid, and — + jz = I ; 

Of tr 



where j = KG; 



y = -V0 2 — x 2 , 



.'. ^JT = -V« 2 - **, JTG = -Vtf 2 - **, 



are the semi-axes of the variable ellipse EGFH, which, by moving along 
OX, generates the ellipsoid. Hence 



hence 



Area EGFH = ir{EK . GK) = ^O 2 - x 2 ) = -X> 



7T^ 

Elementary volume — —^ {a 2 — x 2 )Ax 



x a = 



X a , N , ( a 2 * 2 # 4 ) ' 






— I {a 2 — x 2 )dx 

a fJo 



= i«. 



ivbc C a 
V = — / 2 - ^ 2 )^c = fycabe. 

9. Hemisphere. — Make a = l> = c, and ,# = f a, V = -Jn-tf 3 . 



CENTRE OF GRAVITY OF SYMMETRICAL BODIES. 239 

If the section X were oblique to OX, making an angle 
with it, the elementary volume would not be Xdx, but Xdx sin 0, 

and we should have 

sin 6 fXxdx fXxdx , F , . Aryrj 

sin Of Xdx J Xdx 



3. Oblique Cone (Fig. 151). — Let OA = k; let area of base be 
A, and let the angle made by OX with the base be ; 



X__x>_ 
A~ h* 



*-tr 



Xo = 



-f 



It 



A C h j hl 

— I # 2 a& — 
^Vo 3 



= x? - H 




Fig. 151. 



A C h 

V = sin Of Xdx = ^ sin 6 J x 2 dx = J^ sin 0. 



4. Truncated Cone (Fig. 151). — Let height of entire cone be 
h = OA ; let height of portion cut off be h x ; 



i sin *£ 



— / x*dx 
h?Jk x 


¥ 


4 


J 4 
±h> 

±Ah 


- h* 


•— I x 2 dx 
h 2 J^ 

Ah sin , 
• = r— ( 


fa - 


- ^,3 

3 


sin (A 1 



3# 



240 APPLIED MECHANICS. 



CHAPTER VI. 

STRENGTH OF MATERIALS. 

§ 167. Stress, Strain, and Modulus of Elasticity. — When 
a body is subjected to the action of external forces, if we 
imagine a plane section dividing the body into two parts, the 
force with which one part of the body acts upon the other 
at this plane is called the stress on the plane ; it may be a 
tensile, a compressive, or a shearing stress, or it may be a com- 
bination of either of the two first with the last. In order to 
know the stress completely, we must know its distribution and 
its direction at each point of the plane. If we consider a small 
area lying in this plane, including the point O, and represent 
the stress on this area by p> whereas the area itself is repre- 
sented by a, then will the limit of <- as a approaches, zero be the 

a 

intensity of the stress on the plane under consideration at the 
point O. 

When a body is subjected to the action of external forces, 
and, in consequence of this, undergoes a change of form, it 
will be found that lines drawn within the body are changed, by 
the action of these external forces, in length, in direction, or 
in both ; and the entire change of form of the body may be 
correctly described by describing a sufficient number of these 
changes. 

If we join two points, A and B, of a body before the 
external forces are applied, and find, that, after the application 
of the external forces, the line joining the same two points of 
the body has undergone a change of length A(AB), then is the 



STRESS, STRAIN, AND MODULUS OF ELASTICITY. 241 

limit of the ratio — — -l as AB approaches zero called the 
AB rr 

strain of the body at the point A in the direction AB. 

If AB -h A(AB) > AB, the strain is one of tension. 

If AB + A(AB) < AB, the strain is one of compression. 

Suppose a straight rod of uniform section A to be subjected 
to a pull P in the direction of its length, and that this pull is 
uniformly distributed over the cross-section : then will the in- 
tensity of the stress on the cross-section be 

,-s 

If P be measured in pounds, and A in square inches, then will 
/ be measured in pounds per square inch. 

If the length of the rod before the load is applied be /, 
and its length after the load is applied be I -\- e, then is e the 
elongation of the rod ; and if this elongation is uniform through- 

out the length of the rod, then is - the elongation of the rod 

per unit of length, or the strain. 

Hence, if a represent the strain due to the stress / per 
unit of area, we shall have 

_ e 
a - _. 

The Modulus of Elasticity is commonly defined as the ratio 
of the. stress per unit of area to the strain, or 



and this is expressed in units of weight per unit of area, as in 
pounds per square inch. 

This definition is true, however, only for stresses for which 
Hooke's law " The stress is proportional to the strain " holds. 



24 2 APPLIED MECHANICS. 



For greater stresses the permanent set must first be deducted 
from the strain, and the remainder be used as divisor. 

The limit of elasticity of any material is the stress above 
which the stresses are no longer proportional to the strains. 

The modulus of elasticity was formerly defined as the 
weight that would stretch a rod one square inch in section to 
double its length, if Hooke's law held up to that point, and 
the rod did not break. 

EXAMPLES. 

i. A wrought-iron rod 10 feet long and i inch in diameter is loaded 
in the direction of its length with 8000 lbs. ; find (1) the intensity of 
the stress, (2) the elongation of the rod ; assuming the modulus of the 
iron to be 28000000 lbs. per square inch. 

2. What would be the elongation of a similar rod of cast-iron 
under the same load, assuming the modulus of elasticity of cast-iron to 
be 1 7000000 lbs. per square inch ? 

3. Given a steel bar, area of section being 4 square inches, the 
length of a certain portion under a load of 25000 lbs. being 10 feet, 
and its length under a load of 1 00000 lbs. being io' ©".075 ; find the 
modulus of elasticity of the material. 

4. What load will be required to stretch the rod in the first example 
y 1 ^ inch ? 

§ 168. Resistance to Stretching and Tearing. — The most- 
used criterion of safety against injury for a loaded piece is, 
that the greatest intensity of the stress to which any part of it 
is subjected shall nowhere exceed a certain fixed amount, called 
the working-strength of the material ; this working-strength 
being a certain fraction of the breaking-strength determined 
by practical considerations. 

The more correct but less used criterion is, that the great- 
est strain in any part of the structure shall nowhere exceed 
the working-strain ; the greatest allowable amount of strain 
being a fixed quantity determined by practical considerations. 



RESISTANCE TO STRETCHING AND TEARING. 243 

This is equivalent to limiting the allowable elongation or 
compression to a certain fraction of its length, or the deflection 
of a beam to a certain fraction of the span. 

If the stress on a plane surface be uniformly distributed, 
its resultant will evidently act at the centre of gravity of the 
surface, as has been already shown in § 42 to be the case with 
any uniformly distributed force. 

If a straight rod of uniform section and material be sub- 
jected to a pull in the direction of its length, and if the result- 
ant of the pull acts along a line passing through the centres 
of gravity of the sections of the rod, it is assumed in practice 
that the stress is uniformly distributed throughout the rod, and 
hence that for any section we shall obtain the stress per square 
inch by dividing the total pull by the number of square inches 
in the section. 

If, on the other hand, the resultant of the pull does not act 
through the centres of gravity of the sections, the pull is not 
uniformly distributed ; and while 



/ = _ 

will express the mean stress per square inch, the actual inten- 
sity of the stress will vary at different points of the section, 

P 

being greater than — at some points and less at others. How 

to determine its greatest intensity in such cases will be shown 
later. 

With good workmanship and well-fitting joints, the first 
case, or that of a uniformly distributed stress, can be practi- 
cally realized ; but with ill-fitting joints or poor workmanship, 
or with a material that is not homogeneous, the resultant of 
the pull is liable to be thrown to one side of the line passing 
through the centres of gravity of the sections, and thus there 



244 APPLIED MECHANICS. 

is set up a bending-action in addition to the direct tension, and 
therefore an unevenly distributed stress. 

It is of the greatest importance in practice to take cogni- 
zance of any such irregularities, and determine the greatest 
intensity of the stress to which the piece is subjected : though 
it is too often taken account of merely by means of a factor of 
safety ; in other words, by guess. 

Leaving, then, this latter case until we have studied the 
stresses due to bending, we will confine ourselves to the case 
of the uniformly distributed stress. 

If the total pull on the rod in the direction of its length 

be P, and the area of its cross-section A, we shall have, for the 

intensity of the pull, 

P 



On the other hand, if the working-strength of the material 

per unit of area be f, we shall have, for the greatest admissible 

load to be applied, 

P = /A. 

If / be the working-strength of the material per square 
inch, and E the modulus of elasticity, then is the greatest 
admissible strain equal to 

/ 
a = E 

Thus, assuming 12000 lbs. per square inch as the working 
tensile strength of wrought-iron, and 28000000 lbs. per square 
inch as its modulus of elasticity, its working-strain would be 

12000 3 

a = 



28000000 7000 

Hence the greatest safe elongation of the bar would be 
^__ of its length. Hence a rod 10 feet long could safely be 
stretched yf^ of a foot = 0.05 14". 



VALUES OF BREAKING AND WORKING STRENGTH. 245 

§ 169. Approximate Values of Breaking Strength, and 
of Modulus of Elasticity. — In a later part of this book the 
attempt will be made to give an account of the experiments 
that have been made to determine the strength and elas- 
ticity of the materials ordinarily used in construction, in such 
a way as to enable the student to decide for himself, in any 
special case, upon the proper values of the constants that he 
ought to use. 

For the present, however, the following will be given as a 
rough approximation to some of these quantities, which we may 
make use of in our work until we reach the above-mentioned 
account. 

(a) Cast-iron. 

Breaking tensile strength per square inch, of common quali- 
ties, 14000 to 20000 lbs. ; of gun iron, 30000 to 33000 lbs. 

Modulus of elasticity for tension and for compression, about 
17000000 lbs. per square inch. 

(b) Wrought- Iron. 

Breaking tensile strength per square inch, from 40000 to 
60000 lbs. 

Modulus of elasticity for tension and for compression, about 
28000000. 

(c) Mild Steel. 

Breaking tensile strength per square inch, 55000 to 70000 
lbs. 

Modulus of elasticity for tension and for compression, from 
28000000 to 30000000 lbs. per square inch. 

(d) Wood. 

Breaking compressive strength per square inch : — 

Oak, green 3000 lbs. 

Oak, dry 3000 to 6000 lbs. 

Yellow pine, green 3000 to 4000 lbs. 

Yellow pine, dry 4000 to 7000 lbs. 



246 



APPLIED MECHANICS. 



Modulus of elasticity for compression (average values) : — 

Oak 1300000 lbs. per square inch. 

Yellow pine 1600000 lbs. per square inch. 

§ 170. Sudden Application of the Load. — If a wrought- 
iron rod 10 feet long and 1 square inch in section be loaded 
with 12000 pounds in the direction of its length, and if the 
modulus of elasticity of the iron be 28000000, it will stretch 
0.05 1 4" provided the load be gradually applied : thus, the rod 
begins to stretch as soon as a small load is applied ; and, as the 
load gradually increases, the stretch increases, until it reaches 
0.0514". 

If, on the other hand, the load of 12000 lbs. be suddenly 
applied (i.e., put on all at once) without being allowed to fall 
through any height beforehand, it would cause a greater stretch 
at first, the rod undergoing a series of oscillations, finally 
settling down to an elongation of 0.05 14". 

To ascertain what suddenly applied load will produce at 
most the elongation 0.05 14", observe, that, in the case of the 
gradually applied load, we have a load gradually increasing from 

o to 12000 lbs. 

Its mean value is, therefore, ^(12000) = 6000 lbs. ; and this 
force descends through a distance of 

0.05 1 4". 

Hence the amount of mechanical work done on the rod by the 
gradually applied load in producing this elongation is 

(6000) (0.0514) = 308.4 inch-lbs. 

Hence, if we are to perform upon the rod 308.4 inch-lbs. of 
work with a constant force, and if the stretch is to be 0.05 14", 
the magnitude of the force must be 

3 ° ' 4 ' = 6000 lbs. 
0.05 14' 



RESILIENCE OF A TENSION-BAR. 247 

Hence a suddenly applied load will produce double the strain 
that would be produced by the same load gradually applied ; 
and, moreover, a suddenly applied load should be only half as 
great as one gradually applied if it is to produce the same 
strain. 

§ 171. Resilience of a Tension-Bar. — The resilience of a 
tension-rod is the mechanical work done in stretching it to the 
same amount that it would stretch under the greatest allowable 
gradually applied load, and is found by multiplying the greatest 
allowable load by half the corresponding elongation. 

Thus, suppose a load of 100 lbs. to be dropped upon the 
rod described above in such a way as to cause an elongation not 
greater than 0.05 14", it would be necessary to drop it from a 
height not greater than 3. 08". 

EXAMPLES. 

1. A wrought-iron rod is 12 feet long and 1 inch in diameter, and 
is loaded in the direction of its length; the working-strength of the 
iron being 12000 lbs. per square inch, and the modulus of elasticity 
28000000 lbs. per square inch. 

Find the working-strain. 
Find the working-load. 
Find the working- elongation. 
Find the working-resilience. 

From what height can a 50-pound weight be dropped so as to produce 
tension, without stretching it more than the working- elongation? 

2. Do the same for a cast-iron rod, where the working-strength is 
5000 pounds per square inch, and the modulus of elasticity 17000000; 
the dimensions of the rod being the same. 

§ 172. Results of Wohler's Experiments on Tensile 
Strength. — According to the experiments of Wohler, of which 
an account will be given later, the breaking-strength of a piece 



248 APPLIED MECHANICS. 

depends, not only on whether the load is gradually or suddenly 
applied, but also on the extreme variations of load that the 
piece is called upon to undergo, and the number of changes to 
which it is to be submitted during its life. 

For a piece which is always in tension, he determines the 
following two constants ; viz., /, the carrying-strength per square 
inch, or the greatest quiescent stress that the piece will bear, 
and u, the primitive safe strength, or the greatest stress per 
square inch of which the piece will bear an indefinite number 
of repetitions, the stress being entirely removed in the inter- 
vals. 

This primitive safe strength, u, is used as the breaking- 
strength when the stress varies from o to u every time. Then, 
by means of Launhardt's formula, we are able to determine the 
ultimate strength per square inch for any different limits of 
stress, as for a piece that is to be alternately subjected to 80000 
and 6000 pounds. 

Thus, for Phcenix Company's axle iron, Wohler finds 

t = 3290 kil. per sq. cent. = 46800 lbs. per sq. in., 
u = 2100 kil. per sq. cent. = 30000 lbs. per sq. in. 



Launhardt's formula for the ultimate strength per unit of. area 
is 

t — u least stress ) 
greatest stress)* 



a — u\ 



u 



Hence, with these values of / and u, we should have, for the 
ultimate strength per square inch, 

1 least stress 
a — 2IOO-! 1 + 



( 1 least stress ) 

\ 1 H S-kil. per sq. cent., 

( 2 greatest stress) r 

ii least stress ) 
! _j_ \ ib s# per sq. in. 
2 greatest stress r 



WOHLER'S EXPERIMENTS ON TENSILE STRENGTH. 249 

Thus, if least stress = 6000, and greatest — 80000, we should 

have 

a = 30000$ 1 + \ . -fitl = 30000^1 + -£j\ = 31125; 

if least stress = 60000, and greatest = 80000, 

a = 300005 1 + \ . %\ = 30000^1 4- || = 412505 

if least stress = greatest stress = 80000, 

a = 30000^1 -f \\ = 45000 = carrying-strength. 

Hence, instead of using, as breaking-strength per square inch 
in all cases, 45000, we should use a set of values varying from 
45000 down to 30000, according to the variation of stress which 
the piece is to undergo. 

For working-strength, Weyrauch divides this by 3 : thus 
obtaining, for working-strength per square inch, 

, ( , 1 least stress ) ,, 

= 10000 {1 -\ v lbs. per sq. in. : 

( 2 greatest stress ) 

|.>r Krupp's cast-steel, notwithstanding the fact that Wohler 

finds 

t = 7340 kil. per sq. cent. = 104400 lbs. per sq. in., 
* = 33°° kil. per sq. cent. = 46900 lbs. per sq. in., 

Weyrauch recommends 

( o least stress ) 

a = 3300^ 1 4- — > kil. per sq. cent., 

( 11 greatest stress j ^ H ' 

( o least stress ) 

a = 46900 < 1 4- — Ubs. per sq. m., 

y [ 11 greatest stress ( F H ' 

. ( o least stress ) 

-*- & = 15633 1 1 + r T — ^-7—. — r lbs - P er s q- in - 

l 11 greatest stress j ^ 

EXAMPLES. 

Find the breaking-strength per square inch for a wrought-iron tension 
rod. 

1. Extreme loads are 75000 and 6000 lbs. 

2. Extreme loads are 120000 and 1 00000 lbs. 

3. Extreme loads are 300000 and 10000 lbs. 
Find the safe section for the rod in each case. 



250 APPLIED MECHANICS. 

§173. Suspension-Rod of Uniform Strength. — In the 

case of a long suspension-rod, the weight of the rod itself some- 
times becomes an important item. The upper section must, of 
course, be large enough to bear the weight that is hung from 
the rod plus the weight of the rod itself; but it is sometimes 
desirable to diminish the sections as they descend. This is often 
accomplished in mines by making the rod in sections, each section 
being calculated to bear the weight below it plus its own weight. 
Were the sections gradually diminished, so that each section 
would be just large enough to support the weight below it, we 
should, of course, have a curvilinear form ; and the equation of 
this curve could be found as follows, or, rather, the area of any 
section at a distance from the bottom of the rod. 
Let W = weight hung at O (Fig. 152), 

Let w = weight per unit of volume of 

the rod, 

Let x = distance A O, 

Let 5 = area of section A, 

Let x + dx = distance BO, 



Let 5 + dS = area of section at B, 
Let / = working-strength of the mate- 

rial per square inch. 

i°. The section at O must be just large enough 
to sustain the load IV; 

W 

T 



Fig. 152. •*• *->c 



2°. The area in dS must be just enough to sustain the 
weight of the portion of the rod between A and B. 
The weight of this portion is wSdx ; 

wSdx 



dS = 



7 



dS ■ w , w 

— = — dx .% logeS = yx -f a constant 



CYLINDERS SUBJECTED TO INTERNAL PRESSURE. .25 1 



w 

When x = o, 5 = -^ ; 

1 w U , o , /^\ » 

■\ log^-y = the constant .% log* .5 — log*l -y- 1 =s y# 

= e f /. 5 = ye* • 



- (7) 

This gives us the means of determining the area at any dis- 
tance x from O. 

EXAMPLES. 

1. A wrought-iron tension -rod 200 feet long is to sustain a load of 
2000 lbs. with a factor of safety of 4, and is to be made in 4 sections, 
each 50 feet long; find the diameter of each section, the weight of the 
wrought iron being 480 lbs. per cubic foot. 

2. Find the diameter needed if the rod were made of uniform 
section, also the weight of the extra iron necessary to use in this case. 

3. Find the equation of the longitudinal section of the rod, assum- 
ing a square cross-section, if it were one of uniform strength, instead of 
being made in 4 sections. 

§ 174. Thin Hollow Cylinders subjected to an Internal 
Normal Pressure. — Let/ denote the uniform intensity of the 
pressure exerted by a fluid which is confined within a hollow 
cylinder of radius r and of thickness / (Fig-. 153), e 

the thickness being small compared with the radius, /f ^\ 

Let us consider a unit of length of the cylinder, and c | ~J D 

let us also consider the forces acting on the upper ^z^-^y 
half-ring CED. Pig. 153. 

The total upward force acting on this half-ring, in conse- 
quence of the internal normal pressure, will be the same as 
that acting on a section of the cylinder made by a plane pass- 
ing through its axis, and the diameter CD. The area of this 



252 APPLIED MECHANICS. 

section will be 2r X i = 2r ; hence the total upward force will be 
2r X / = 2pr; and the tendency of this upward force is to cause 
the cylinder to give way at A and B, the upper part separating 
from the lower. 

This tendency is resisted by the tension in the metal at the 
sections AC and BD ; hence at each of these sections, there has 
to be resisted a tensile stress equal to \{2pr) =pr. This stress 
is really not distributed uniformly throughout the cross-section 
of the metal ; but, inasmuch as the metal is thin, no serious 
error will be made if it be accounted as distributed uniformly. 
The area of each section, however, is t X i =*; therefore, if 
T denote the intensity of the tension in the metal in a tangential 
direction (i.e., the intensity of the hoop tension), we shall have 

t 

Hence, to insure safety, T must not be greater than /, the 
working-strength of the material for tension ; hence, putting 

we shall have 

'-5 

as the proper thickness, when/ = normal pressure per square 
inch, and radius = r. 

The above are the formulae in common use for the deter- 
mination of the thickness of the shell of a steam-boiler ; for in 
that case the steam-pressure is so great that the tension 
induced by any shocks that are likely to occur, or by the weight 
of the boiler, is very small in comparison with that induced 
by the steam-pressure. On the other hand, in the case of an 
ordinary water-pipe, the reverse is the case. 



RESISTANCE TO DIRECT COMPRESSION. <2$$ 

To provide for this case, Weisbach directs us to add to the 
thickness we should obtain by the above formulae, a constant 
minimum thickness. 

The following are his formulae, d being the diameter in 
inches, / the internal normal pressure in atmospheres, and / 
the thickness in inches. For tubes made of 

Sheet-iron t= 0.00086/*/ -f- 0.12 

Cast-iron / = 0.00238 pd -f- 0.34 

Copper / = 0.00148/*/ -f- 0.16 

Lead / = 0.00507/*/ + 0.21 

Zinc t = 0.00242/*/ + 0.16 

Wood t — 0.03230/*/ + 1.07 

Natural stone / = 0.03690/*/ 4- 1.18 

Artificial stone / = 0.05380/*/+ 1.58 

§ 175. Resistance to Direct Compression. — When a piece 
is subjected to compression, the distribution of the compressive 
stress on any cross-section depends, first, upon whether the 
resultant of the pressure acts along the line containing the cen- 
tres of gravity of the sections, and, secondly, upon the dimen- 
sions of the piece ; thus determining whether it will bend or 
not. 

In the case of an eccentric load, or of a piece of such length 
that it yields by bending, the stress is not uniformly distributed ; 
and, in order to proportion the piece, we must determine the 
greatest intensity of the stress upon it, and so proportion it 
that this shall be kept within the working-strength of the ma- 
terial for compression. 

Either of these cases is not a case of direct compres- 
sion. 

In the case of direct compression (i.e., where the stress over 
each section is uniformly distributed), the intensity of the stress 
is found by dividing the total compression by the area of the 



2 54 APPLIED MECHANICS. 

section ; so that, if P be the total compression, and A the area 
of the section, and/ the intensity of the compressive stress, 

On the other hand, if f is the compressive working-strength ol 
the material per square inch, and A the area of the section in 
square inches, then the greatest allowable load on the piece 
subjected to compression is 

P=fA. 

The same remarks as were made in regard to a suddenly 
applied load and resilience, in the case of direct tension, apply 
in the case of direct compression. 

§ 176. Results of Wohler's Experiments on Compressive 
Strength. — Wohler also made experiments in regard to pieces 
subjected to alternate tension and compression, taking, in the 
experiments themselves, the case where the metal is subjected 
to alternate tensions and compressions of equal amount. 

The greatest stress of which the piece would bear an indefi- 
nite number of changes under these conditions, is called the 
vibration safe strength, and is denoted by s. 

Weyrauch deduces a formula similar to that of Launhardt 
for the greatest allowable stress per unit of area on the piece 
when it is subjected to alternate tensions and compressions of 
different amounts. 

Thus, for Phoenix Company's axle iron, Wohler deduces 

/ = 3290 kil. per sq. cent. = 46800 lbs. per sq. in., 
u — 2100 kil. per sq. cent. = 30000 lbs. per sq. in., 
s = 1 1 70 kil. per sq. cent. = 16600 lbs. per sq. in. 



EXPERIMENTS ON COMPRESSIVE STRENGTH. 255 

«=> — ■ — ■ • 

Weyrauch's formula for the ultimate strength per unit of 
area is 

)u — s least maximum stress ) 
1 " t~t : : 1> 
u greatest maximum stress \ 

and, with these values of u and s, it gives 



1 least maximum stress 

2 greatest maximum stress 

or 

least maximum stress 



a = 2100 { 1 — ^ — > kil. per sq. cent., 



(1 least maximum stress ) 
1 — — — : - > lbs. per sq. in. 
2 greatest maximum stress J ' ■ 



With a factor of safety of 3, we should have, for the greatest 
admissible stress per square inch, 



b = 1 0000 



{1 least maximum stress \ 
2 greatest maximum stress j 



For Krupp's cast-steel, 

t = 7340 kil. per sq. cent. = 104400 lbs. per sq. in., 

u = 3300 kil. per sq. cent. = 46900 lbs. per sq. in. approximately, 

s = 2050 kil. per sq. cent. = 29150 lbs. per sq. in. approximately. 

We have, therefore, for the breaking-strength per unit of 
area, according to Weyrauch's formula, 

least maximum stress 



{e least maximum stress ) 
1 — - — — = > kil. per sq. cent., 
1 1 greatest maximum stress ) r ^ 

( c least maximum stress ) 

a — 46900 \ 1 — — — : J. lbs. per sq. in. ; 

( JI greatest maximum stress! r n 



2^6 APPLIED MECHANICS. 

and, using a factor of safety of 3, we have, for the greatest admis* 
sible stress per square inch, 

( e least maximum stress ) 

b — 15630 { 1 — f- — - : >lbs. per sq. in. 

{ ll greatest maximum stress) r n 

The principles respecting an eccentric compressive load, and 
those respecting the giving-way of long columns so far as they 
are known, can only be treated after we have studied the resist- 
ance of beams to bending; hence this subject will be deferred 
until that time. 

EXAMPLES. 

Find the proper working and breaking strength per square inch to 
be used for a wrought- iron rod, the extreme stresses being — 

1. 80000 lbs. tension and 6000 lbs. compression. 

2. 1 00000 lbs. tension and 1 00000 lbs. compression. 

3. 70000 lbs. tension and 60000 lbs. compression. 
Do the same for a steel rod. 

§ 177. Resistance to Shearing. — One of the principal cases 
where the resistance to shearing comes into practical use is 
that where the members of a structure, which are themselves 
subjected to direct tension or compression or bending, are united 
by such pieces as bolts, rivets, pins, or keys, which are sub- 
jected to shearing. Sometimes the shearing is combined with 
tension or with bending ; and whenever this is the case, it is 
necessary to take account of this fact in designing the pieces. 
It is important that the pins, keys, etc., should be equally 
strong with the pieces they connect. 

Probably one of the most important modes of connection is 
by means of rivets. In order that there may be only a shearing 
action, with but little bending of the rivets, the latter must 
fit very tightly. The manner in which the riveting is done will 
necessarily affect very essentially the strength of the joints; 



RESISTANCE TO SHEARING. 



257 



hence the only way to discuss fully the strength of riveted 
joints is to take into account the manner of effecting the rivet- 
ing, and hence the results of experiments. These will be 
spoken of later ; but the ordinary theories by which the strength 
and proportions of .some of the simplest forms of riveted joints 
are determined will be given, which theories are necessary also 
in discussing the results of experiments thereon. 

The principle on which the theory is based, in these simple 
cases, is that of making the resistance of the joint to yielding 
equal in the first three, and also in either the fourth or the 
fifth of the ways in which it is possible for it to yield, as 
enumerated on pages 258 and 259. 



A single-riveted lap-joint is one 
with a single row of rivets, as 
shown in Fig. 154. 



A single-riveted butt-joint with 
one covering plate is shown in 
Fig. 155. 




Fig. 



Fig. 155. 



A single-riveted butt-joint with 
two covering plates is shown in 
156. 



Fig. 



_J 

Fig. 156. 



258 



APPLIED MECHANICS. 




Fig. 157. 




A double-riveted lap-joint with 
the rivets staggered is shown in 
Fig- I 57> one with chain riveting, 
in Fig. 158. 



Fig. 158. 



Taking the case of the single-riveted lap-joint shown in Fig. 
1 54, it may yield in one of five ways : — 




i°. By the crushing of the plate 
in front of the rivet (Fig. 159). 



Fig. 159. 



Fig. 160. 



3 2°. By the shearing of tne rivet 

(Fig. 160). 



RESISTANCE TO SHEARING. 



259 




Fig. 161. 



centre 



3 . By the tearing of the plate 
between the rivet-holes (Fig. 161). 

4 . By the rivet breaking 
through the plate (Fig. 162). 

5 . By the rivet shearing out 
the plate in front of it. 



Let us call 

d the diameter of a rivet. 

/ the pitch of the rivets ; i.e., FlG - l6z - 

their distance apart from centre to centre. 
t the thickness of the plate. 
/ ihe lap of the plate ; i e., the distance from the 

of a livet-hole to the outer edge of the plate. 

2 
f t the ultimate tensile strength of the iron. 
f s the ultimate shearing-strength of the rivet-iron. 
fs the ultimate shearing-strength of the plate. 
f e the ultimate crushing-strength of the iron. 
We shall then have — 

i°. Resistance of plate in front of rivet to crushing = f c td. 

2°. Resistance of one rivet to shearing = fi- — V 

3 . Resistance of plate between two rivet-holes to tearing 
= ZAP- d). 

4 . Resistance of plate to being broken through 

where a is a constant depending on the material. This may be 
taken as empirical for the present. 

A reasonable value of this constant is ^f t . 



= a — , 
d 



260 APPLIED MECHANICS. 

5°. Resistance of plate in front of the rivet to shearing 

= 2f s dt. 

Assuming that we know the thickness of the plate to start 
with, we obtain, by equating the first two resistances, 

fctd = fs— .'. d = — j, 

which determines the diameter of the rivet. 
Equating 3 and 2 , we obtain 

which gives the pitch of the rivets in terms of the diameter of 
the rivet, and the thickness of the plate. 
Equating, next, 4 and i°, we have 

a%=f c td .-. h=d\T-> 
a T a 

which gives the lap of the plate needed in order that it may not 
break through. 

By equating 5 and i°, we find the lap needed that it may 
not shear out in front of the rivet. 

A similar method of reasoning would enable us to determine 
the corresponding quantities in the cases of double-riveted 
joints, etc. 

There are a number of practical considerations which 
modify more or less the results of such calculations, and which 
can only be determined experimentally. A fuller account of 
this subject from an experimental point of view will be given 
later. 

§ 178. Intensity of Stress. — Whenever the stress over a 
plane area is uniformly distributed, we obtain its intensity at 
each point by dividing the total stress by the area over which 
it acts, thus obtaining the amount per unit of area. When, how- 
ever, the stress is not uniformly distributed, or when its inten- 



INTENSITY OF STRESS. 



26l 



sity varies at different points, we must adopt a somewhat differ- 
ent definition of its intensity at a given point. In that case, if 
we assume a small area containing that point, and divide the 
stress which acts on that area by the area, we shall have, in the 
quotient, an approximation to the intensity required, which will 
approach nearer and nearer to the true value of the intensity at 
that point, the smaller the area is taken. 

Hence the intensity of a variable stress at a given point is, — 

The limit of the ratio of the stress acting on a small area 
containing that point, to the area, as the latter grows smaller and 
smaller. 

By dividing the total stress acting on a certain area by the 
entire area, we obtain the mean intensity of the stress for the 
entire area. 

§ 179. Graphical Representation of Stress. — A conven- 
ient mode of representing stress 
graphically is the following: — 

Let AB (Fig. 163) be the plane 
surface upon which the stress acts ; 
let the axes OX and OY be taken 
in this plane, the axis OZ being at 
right angles to the plane. 

Conceive a portion of a cylinder 
whose elements are all parallel to 
OZ, bounded at one end by the 
given plane surface, and at the 
other by a surface whose ordinate at any point contains as 
many units of length as there are units of force in the intensity 
of the stress at that point of the given plane surface where the 
ordinate cuts it. 

The volume of such a figure will evidently be 

V = ffzdxdy = ffpdxdy, 




Fig. 163. 



where z ~ p ~ intensity of the stress at the given point. 



262 



APPLIED MECHANICS. 




Hence the volume of the cylindrical figure will contain as 
many units of volume as the total stress contains units of 
force ; or, in other words, the total stress will be correctly repre- 
sented by the volume of the body. 

If the stress on the plane 
figure is partly tension and 
partly compression, the sur- 
face whose ordinates repre- 
sent the intensity of the 
stress will lie partly on one 
side of the given plane sur- 
face and partly on the other ; 
this surface and the plane 
surface on which the stress 
acts, cutting each other in 
some line, straight or curved, 
as shown in Fig. 164. In that 
case, the magnitude of the resultant stress P = V = ffzdxdy 
will be equal to the difference of the wedge-shaped volumes 
shown in the figure. 

It will be observed that the above method of representing 
stress graphically represents, i°, the intensity at each point of 
the surface to which it is applied ; and, 2°, the total amount 
of the stress on the surface. It does not, however, represent 
its direction, except in the case when the stress is normal to 
the surface on which it acts. 

In this latter case, however, this is a complete representa- 
tion of the stress. 

The two most common cases of stress are, i°, uniform stress, 
and, 2°, uniformly varying stress. These two cases are repre- 
sented respectively in Figs. 165 and 166; the direction also 
being correctly represented when, as is most frequently the 
case, the stress is normal to the surface of action. In Fig. 
165, AB is supposed to be the surface on which the stress 



Fig. 164. 



GRAPHICAL REPRESENTATION OF STRESS. 



263 





acts ; the stress is supposed to be uniform, and normal to the 

surface on which it acts ; the bound- 
ing surface in this case becomes a 

plane parallel to AB; the intensity 

of the stress at any point, as P, will 

be represented by PQ; while the 

whole cylinder will contain as many 

units of volume as there are units of 

force in the whole stress. 

Fig. 166 represents a uniformly 

varying stress. Here, again, AB is 

the surface of action, and the stress 

is supposed to vary at a uniform rate 

from the axis Y The upper bounding surface of the cylin- 
drical figure which represents the stress 
becomes a plane inclined to the XOY 
plane, and containing the axis O Y. 

In this case, if a represent the in- 
tensity of the stress at a unit's distance 
from OY, the stress at a distance x from 
OY will hep — ax, and the total amount 
of the stress will be 



Fig. 165. 




Fig. 166. 



P = ffpdxdy — affxdxdy. 



When a stress is oblique to the surface of action, it may be 
represented correctly in all particulars, except in direction, in 
the above-stated way. 

§180. Centre of Stress. — The centre of stress, or the 
point of the surface at which the resultant of the stress acts, 
often becomes a matter of practical importance. If, for con- 
venience, we employ a system of rectangular co-ordinate axes, 
of which the axes OX and OY are taken in the plane of the 
surface on which the stress acts, and if we let / = <f>(x, y) be 



264 APPLIED MECHANICS. 

the intensity of the stress at the point (x, y), we shall have, 
for the co-ordinates of the centre of stress, 

_ ffxpdxdy __ ffypdxdy 

ffpdxdy ' l ffpdxdy ' 

(see § 42), where the denominator, or ffpdxdy, represents the 
total amount of the stress. 

When the stress is positive and negative at different parts 
of the surface, as in Fig. 164, the case may arise when the posi- 
tive and negative parts balance each other, and hence the 
stress on the surface constitutes a statical couple. In that case 

ffpdxdy = o. 

§ 181. Uniform Stress. — In the case of uniform stress, we 
have — 

i°. The intensity of the stress is constant, or / = a con- 
stant. 

2°. The volume which represents it graphically becomes a 
cylinder with parallel and equal bases, as in Fig. 165. 

3 . The centre of stress is at the centre of gravity of the 
surface of action ; for the formulae become, when / is constant,- 

pffxdxdy _ ffxdxdy _ 
Xl = pffdxdy = ffdxdy ~ Xo> 

_ pffydxdy _ ffydxdy _ 
y * " = Pffdxdy ~ ffdxdy ~ y °' 

where x ot y Q1 are the co-ordinates of the centre of gravity of the 
surface. 

Examples of uniform stress have already been given in the 
cases of direct tension, direct compression, and, in the case of 
riveted joints, for the shearing-force on the rivet. 



UNIFORMLY VARYING STRESS. 26$ 

§182. Uniformly Varying Stress. — Uniformly varying 
stress has already been denned as a stress whose intensity varies 
uniformly from a given line in its own plane ; and this line will 
be called the Neutral Axis. Thus, if the plane be taken as the 
XOY plane (Fig. 166), and the given line be taken as OY, we 
shall have, if a denotes the intensity of the stress at a unit's 
distance from O Y, and x the distance of any special point from 
O Y, that the intensity of the stress at the point will be 

p = ax. 

The total amount of the stress will be 

P — affxdxdy. 

The total moment of the stress about O Y will be found by 
multiplying each elementary stress by its leverage. This lever- 
age is, in the case of normal stress, x ; hence in that case the 
moment of any single elementary force will be 

(ax Ax Ay) x = ax* Ax Ay, 
and the total moment of the stress will be 

M = affx^dxdy = al. 

In the case of oblique stress, this result has to be modified, 
as the leverage is no longer x. Confining ourselves to stress 
normal to the plane of action, we have, for the co-ordinates of 
the centre of stress, 

SSpxdxdy affx 2 dxdy _ ffx 2 dxdy _ ffx 2 dxdy_ I 
Jfpdxdy P ffxdxdy x A x A 

Sfpydxdy affxydxdy ffxydxdy ffxydxdy 



since 



Sfpdxdy P ffxdxdy XoA 

P = affxdxdy = ax^A, 



where x m y , are the co-ordinates of the centre of gravity, and 
A is the area of the surface of action. 



266 



APPLIED MECHANICS. 



§ 183. Case of a Uniformly Varying Stress which 
amounts to a Statical Couple. — Whenever P = o, we have 



af fxdxdy = o 



ffxdxdy = o 



x A = o 



x n = o. 



In this case, therefore, we have — 

i°. There is no resultant stress, and hence the whole stress 
amounts to a statical couple. 

2°. Since x Q = o, the centre of gravity of the surface of 
action is on the axis OY, which is the neutral axis. 

Hence follows the proposition : — 

When a uniformly varying stress amounts to a statical couple, 
the neutral axis contains {passes through) the centre of gravity 
of the surface of action. 

In this case there is no single resultant of the stress ; but 
the moment of the couple will be, as has been already shown, 

M = affx 2 dxdy. 

§ 184. Example of Uniformly Varying Stress One of 

the most common examples of uniformly varying stress is that 
of the pressure of water upon the sides of the vessel contain- 



Thus, let Fig. 167 represent the vertical cross-section of a 
reservoir wall, the water pressing against the 
vertical face AB. It is a fact established by 
experiment, that the intensity of the pressure 
of any body of water at any point is propor- 
tional to the depth of the point below the 
free upper level of the water, and normal to 
the surface pressed upon. Hence, if we sup- 
pose the free upper level of the water to be 
even with the top of the wall, the intensity 
of the pressure there will be zero ; and if we represent by CB 
the intensity of the pressure at the bottom, then, joining^ and 





1 \ 




\ 




1 \ 




\ 




1 \ 




]\ 




\ 


/e d 


1 I l\ 








1 1 \ 








1 1 \ 






/ B 


1 ' \ 


1 l\ 


c 





Fig. 167. 



STRESSES IN BEAMS UNDER TRANSVERSE LOAD. 267 

C we shall have the intensity of the pressure at any point, as 
D } represented by ED, where 

ED : CB = AD: AB. 



Here, then, we have a case of uniformly varying stress nor- 
mal to the surface on which it acts. 

§ 185. Fundamental Principles of the Common Theory 
of the Stresses in Beams under 
a Transverse Load. — Fig. 168 
shows a beam fixed at one end and 
loaded at the other, while Fig. 169 
shows a beam supported at the 
ends and loaded at the middle. 
Let, in each case, the plane of the 
paper contain a vertical longi- 
tudinal section of the beam. In 
o Fig. 168, 
}\ it is evi- 
dent that 
I the upper 



I 




I 



fibres are lengthened, while the lower 
ones are shortened, and vice versa in 
Fig. 169. In either case, there is, 
somewhere between the upper and 
lower fibres, a fibre which is neither 
elongated nor com- 
pressed. 

Let CN repre- 
sent that fibre, Fig. 

168, and CP, Fig. 

169. This line may 
be called the neutral 
line of the longitu- 
dinal section ; and, if a section be made at any point at right 




Fig. 169. 



268 APPLIED MECHANICS. 

angles to this line, the horizontal line which lies in the cross- 
section, and cuts the neutral lines of all the longitudinal sec- 
tions, or, in other words, the locus of the points where the 
neutral lines of the longitudinal sections cut the cross-section, 
is called the Neutral Axis of the cross-section. In the ordinary- 
theory of the stresses in beams, a number of assumptions are 
made, which will now be enumerated. 

ASSUMPTIONS MADE IN THE COMMON THEORY OF BEAMS. 

Assumption No. i. — If, when a beam is not loaded, a 
plane cross-section be made, this cross-section will still be a 
plane after the load is put on, and bending takes place. From 
this assumption, we deduce, as a consequence, that, if a certain 
cross-section be assumed, the elongation or shortening per unit 
of length of any fibre at the point where it cuts this cross-sec- 
tion, is proportional to the distance of the fibre from the neutral 
axis of the cross-section. 

Proof. — Imagine two originally parallel cross-sections so 
near to each other that the curve in which that part of the 
neutral line between them bends may, without appreciable error, 
be accounted circular. Let ED and GH (Fig. 168 or Fig. 169) 
be the lines in which these cross-sections cut the plane of the 
paper, and let be the point of intersection of the lines ED 
and GH. Let OF — r, FL = y, FK = /, LM =/+«/, in 
which a is the strain or elongation per unit of length of a fibre 
at a distance y from the neutral line, y being a variable ; then, 
because FK and LM are concentric arcs subtending the same 
angle at the centre, we shall have the proportion 



r -f y l 4- <rf y 

= — -. — or 1 + a = 1 H — > 

r I r 



y 

a = - or 

r 



&> 



ASSUMPTIONS IN THE COMMON THEORY OF BEAMS. 269 

but as y varies for different points in any given cross-section, 
while r remains the same for the same section, it follows, that, 
if a certain cross-section be assumed, the strain of any fibre at 
the point where it cuts this cross-section is proportional directly 
to the distance of this fibre from the neutral axis of the cross- 
section. 



Assumption No. 2. — This assumption is that commonly 
known as Hooke s Law. It is as follows : " Ut tensio sic vis ; " 
i.e., The stress is proportional" to the strain, or to the elonga- 
tion or compression per unit of length. As to the evidence in 
favor of this law, experiment shows, that, as long as the mate- 
rial is not strained beyond safe limits, this law holds. Hence, 
making these two assumptions, we shall have : At a given 
cross-section of a loaded beam, the direct stress on any fibre 
varies directly as the distance of the fibre from the neutral axis. 
Hence it is a uniformly varying stress, and we may repre- 
sent it graphically as follows : Let 
ABCD, Fig. 170, be the cross-sec- 
tion of a beam, and KL the neutral 
axis. Assume this for axis OY, and 
draw the other two axes, as in the 
figure. If, now, EA be drawn to 
represent the intensity of the direct 
(normal) stress at A, then will the 
pair .of wedges AEFBKL and FlG " I7 °- 

DCHGKL represent the stress graphically, since it is uni- 
formly varying. 




POSITION OF NEUTRAL AXIS. 



ASSUMPTION No. 3. — It will next be shown that, on the 
two assumptions made above, and from the further assumption 
that the deformation of each fibre of the beam parallel to its 
longitudinal axis is due to the forces acting on its ends alone 



2/0 



APPLIED MECHANICS. 



and that it suffers no traction from neighboring fibres, it fol- 
lows that the neutral axis must pass through the centre of 
gravity of the cross-section. 




Fig. 171. 



S 



N 
Fig. 172. 



Since the curvatures in Figs. 168 and 169 are exaggerated 
in order to render them visible, Figs. 171 and 172 have been 
drawn. If, now, we assume a section DE, such that AD = x 
(Fig. 171) and NE = x (Fig. 172), and consider all the forces 
acting on that part of the beam which lies to -the right of DE 
(i.e., both the external forces and the stresses which the other 
parts of the beam exert on this part), we must find them in 
equilibrium. The external forces are, in Fig. 172, — 

i°. The loads acting between B and E ; in this case there 
are none. 

2 . The supporting force at B ; in this case it is equal to 

— , and acts vertically upwards. 

In Fig. 171 they are, — 

The loads between D and N ; in this case there is only the 
one, W at N. 

The internal forces are merely the stresses exerted by the 
other parts of the beam on this part : they are,- — 

i°. The resistance to shearing at the section, which is a 
vertical stress. 

2 . The direct stresses, which are horizontal. 

Now, since the part of the beam to the right of DE is at 
rest, the forces acting on it must be in equilibrium ; and, since 



POSITION OF NEUTRAL AXIS. 2 7 l 

they are all parallel to the plane of the paper, we must have 
the three following conditions ; viz., — 

i°. The algebraic sum of the vertical forces must be zero. 

2°. The algebraic sum of the horizontal forces must be zero. 

3°. The algebraic sum of the moments of the forces about 
any axis perpendicular to the plane of the paper must be 
zero. 

But, on the above assumptions, the only horizontal forces 
are the direct stresses : hence the algebraic sum of these direct 
stresses must be zero ; or, in other words, the direct stresses 
must be equivalent to a statical couple. 

Now, it has already been shown, that, whenever a uniformly 
varying stress amounts to a statical couple, the neutral axis 
must pass through the centre of gravity of the surface acted 
upon. Hence in a loaded beam, if the three preceding assump- 
tions be made, it follows that the neutral axis of any cross- 
section must contain the centre of gravity of that section. 

By way of experimental proof of this conclusion, Barlow 
has shown by experiment, that, in a cast-iron beam of rectangu- 
lar section, the neutral axis does pass through the centre of 
gravity of the section. 

RESUME. 

The conclusions arrived at from the foregoing are as fol- 
lows : — 

i°. That at any section of a loaded beam, if a horizontal 
line be drawn through the centre of gravity of the section, 
then the fibres lying along this line will be subjected neither 
to tension nor to compression ; in other words, this line will be 
the neutral axis of the section. 

2°. The fibres on one side of this line will be subjected to 
tension, those on the other side being subjected to compres- 
sion ; the tension or compression of any one fibre being propor- 
tional to its distance from the neutral axis. 



272 APPLIED MECHANICS. 

The first of the three assumptions of the common theory 
was not accepted by St. Venant, who developed by means of 
the methods of the Theory of Elasticity a theory of beams 
based upon the second and third assumptions only. A study 
of St. Venant's theory involves, however, far more complica- 
tion, and requires a good previous knowledge of the Theory of 
Elasticity. Moreover the results of the two theories as far as 
the determination of the outside fibre-stresses and of the de- 
flections are practically in agreement, while, on the other hand, 
the intensities of the shearing-forces as computed by the two 
theories are not in agreement. 

The St. Venant theory may be found in several treatises 
upon the Theory of Elasticity. 

§ 186. Shearing-Force and Bending-Moment. — In deter- 
mining the strength of a beam, or the proper dimensions of a 
beam to bear a certain load, when we assume the neutral axis 
to pass through the centre of gravity of the cross-section, we 
have imposed the second of the three last-mentioned conditions 
of equilibrium. The remaining two conditions may otherwise 
be stated as follows : — 

i°. The total force tending to cause that part of the beam 
that lies to one side of the section to slide by the other part, 
must be balanced by the resistance of the beam to shearing at 
the section. 

2 . The resultant moment of the external forces acting on 
that part of the beam that lies to one side of the section, about 
a horizontal axis in the plane of the section, must be balanced 
by the moment of the couple formed by the resisting stresses. 

The shearing-force at any section is the force with which the 
part of the beam on one side of the section tends to slide by the 
part on the other side. In a beam free at one end, it is equal to 
the sum of the loads between the section and the free end. In 
a beam supported at both ends, it is equal in magnitude to the 
difference between the supporting force at either end, and 
the sum of the loads between the section and that support. 



SHEARING-FORCE AND BENDING-MOMENT. 273 



The bending-moment at any section is the resultant moment 
of the external forces acting on the part of the beam to one 
side of the section, these moments being taken about a hori- 
zontal axis in the section. 

In a beam free 'at one end, it is equal to the sum of the 
moments of the loads between the section and the free end, 
about a horizontal axis in the section. 

In a beam supported at both ends, it is the difference be- 
tween the moment of either supporting force, and the sum of 
the moments of the loads between the section and that sup- 
port ; all the moments being taken about a horizontal axis in 
the section. 

Hence the two conditions of equilibrium may be more 
briefly stated as follows : — 

i°. The shearing-force at the section must be balanced 
by the resistance opposed by the beam to shearing at the 
section. 

2 . The bending-moment at the section must be balanced 
by the moment of the couple formed by the resisting stresses. 

It is necessary, therefore, in determining the strength of a 
beam, to be able to determine the shearing-force and bending- 
moment at any point, and also the greatest shearing-force and 
the greatest bending-moment, whatever be the loads. 

A table of these values for a number of ordinary cases will 
now be given ; but I should recommend that the table be merely 
considered as a set of examples, and that the rules already 
given for finding them be followed in each individual case. 

Let, in each case, the length of the beam be /, and the 
total load W. When the beam is fixed at one end and free at 
the other, let the origin be taken at the fixed end ; when it is 
supported at both ends, let it be taken directly over one support. 
Let x be the distance of any section from the origin. Then we 
shall have the results given in the following table : — 



274 



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MOMENTS OF INERTIA OF SECTIONS. 275 

In a beam fixed at one end and free at the other, the great- 
est shearing-force, and also the greatest bending-moment, are at 
the fixed end. In a beam supported at both ends, and loaded 
at the middle, or with a uniformly distributed load, the greatest 
shearing-force is at either support, the greatest bending-moment 
being at the middle. In the last case (i.e., that of a beam sup- 
ported at the ends, and having a single load not at the middle), 
the greatest bending-moment is at the load ; the greatest shear- 
ing-force being at that support where the supporting force is 
greatest. 

§ 187. Moments of Inertia of Sections. — In the usual 
methods of determining the strength of a beam or column, it 
is necessary to know, i°, the distance from the neutral axis of 
the section to the most strained fibres ; 2°, the moment of in- 
ertia of the section about the neutral axis. The manner of 
finding the moments of inertia has been explained in Chap. II. 

In the following table are given the areas of a large number 
of sections, and also their moments of inertia about the neutral 
axis, which is the axis YY in each case. These results should 
be deduced by the student. 



276 



APPLIED MECHANICS. 






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MOMENTS OF INERTIA OF SECTIONS, 



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MOMENTS OF INERTIA OF SECTIONS. 



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286 



APPLIED MECHANICS. 



§ 188. Cross-Sections of Phoenix Columns considered 
as made of Lines. — It is to be observed that the moments 
of inertia are the same for all axes passing through the centre. 
Thickness = t, radius of round ones = r t area of each flange 
= a, length of each flange = /. 



Figure. 



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Description. 



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REPRESENTATION OF B ENDING-MOMENTS. 



287 




Fig. 215. 



§ 189. Graphical Representation of Bending-Moments. — 

The bending-moment at each point of a loaded beam may be 
represented graphically by lines laid off to scale, as will be 
shown by examples. 

I. Suppose we have the cantilever shown in Fig. 215, 
loaded at D with a load W: then 
will the bending-moment at any 
section, as at F, be obtained by 
multiplying W by FD ; that at AC 
being W X (AB). If, now, we lay 
off CE to scale to represent this, 
i.e., having as many units of length 
as there are units of moment in the product WX (AB), and 
join E with D, then will the ordinate FG of any point, as G, 
represent (to the same scale) the bending-moment at a section 
through F 

II. If we have a uniformly distributed load, we should have, 
for the line corresponding to CE in Fig. 215, a curve. This is 

shown in Fig. 216, where we have the 
uniformly distributed load EIGF. If 
we take the origin at D, as before, 
we have, for the bending-moment, at a 
distance x from the origin, as has been 

x) 2 ; and by giving x dif- 



TT 



"r^g^ 



Fig. 216. 



w 

shown, — (/ 
2/ 



ferent values, and laying off the corresponding value of the 
bending-moment, we obtain the curve CA, any ordinate of 
which will represent the bending-moment at the corresponding 
point of the beam. 

When we have more than one load on a beam, we must draw 
the curve of bending-moments for each load separately, and 
then find the actual bending-moment at any point of the beam 



283 



APPLIED MECHANICS. 



by taking the sum of the ordinates (drawn from that point) of 
each of these separate curves or straight lines. If we then 
draw a new curve, whose ordinates are these sums, we shall have 
the actual curve of bending-moments for the beam as loaded. 
Some examples will now be given, which will explain them- 
selves. 

III. Fig. 217 shows a cantilever with three concentrated 
loads. The line of bending-moments 
for the load at C is CE, that for the 
load at O is OF, and for the load at P 
is PG. They are combined above the 
beam by laying off AH = DE, HK — 
DF, and KL = DG, and thus obtaining 
the broken line LMNB, which is the 
line of bending-moments of the beam 
loaded with all three loads. 



Fig. 217. 

IV. Fig. 218 shows the case of a beam supported at both 
ends, and loaded at a single point 
D ; ALB is the line of bending- 
moments when the weight of the 
beam is disregarded, so that xy = 
bending-moment at x. FlG> 2l8 

V. Fig. 219 shows the case of a beam supported at the ends, 
and loaded with three concentrated 
loads at the points B, C, and D re- 
spectively ; the lines of bending-mo- 
ments for each individual load being 
respectively AFE, AGE, and AHE y 

fig. 219. and the actual line of bending-mo- 

ments being AKLME. 






REPRESENTATION OF B ENDING-MOMENTS. 



289 




VI. Fig. 220 shows the case of a beam supported at the 
ends, and loaded with a uniformly dis- A 
tributed load ; the line of bending- 
moments being a curve, ACDB t as 
shown in the figure. 

VII. In Fig. 221 we have the case of a beam, over a part of 
which, viz., EF } there is a distributed load ; the rest of the 

beam being unloaded. The line of 
bending-moments is curvilinear be- 
tween E and F f and straight outside 
of these limits. It isAGSHB; and, 
when the curve is plotted, we can 
find the greatest bending-moment 
graphically by finding its greatest ordinate. We can also 
determine it analytically * by first determining the bending- 
moment at a distance x from the origin, and on the side 
towards the resultant of the load, and then differentiating. 
This process is shown in the following: — 
Let A (Fig. 222) be the point where 
the resultant of the load acts, and O the 
middle of the beam, and let w be the 
load per unit of length ; let OA = a, AB = 
AC= b, and ED 




Fig. 221. 



E h 'i ■ 1 I d 

~* 1 h 1 A - 



B 



Fig. 222. 

2c, so that the whole load = 2wb : there- 



fore supporting force at D = 2ivb 



a 



c wb(a + c) 



2C C 

If we take a section at a distance x from O to the right, we 
shall have, for the bending-moment at that section, 

wb(a + c) w . 

(c — x) (a 4- — xy = a maximum. 

c v ' 2 v ' 



Differentiate, and we have 
— wb(a -f- c) 



-f- w{a -\-b — x) = o 



a(c — b) . 
x = 9 



290 



APPLIED MECHANICS. 



hence the greatest bending-moment will be 



wb{a -}- c) / a(c — b\ w. 



ab\* 



/ au — o\ wf , , , ab\ 

IJDu 7£}b 2 

= — (0 + *") (c 2 — ac -\- ab) (a 2 H- 2ac -f- c 2 ) 

c 2 2c 2 

wb , 



= — (a 2 b -~ 2«V -f- 2C* — &*). 

2<T 2 



VIII. In Figs. 223 and 224 we have the case of a beam 
supported at 
the ends, and 
loaded with a 
uniformly dis- 
tributed load, 
and also with 
a concen- 
trated load. 
In the first 
figure, the greatest bending-moment 




Fig. 223. 




Fig. 224. 



is atZ>, and in the second at £ 



IX. In Fig. 225 we have a beam supported at A and B, and 
loaded at C and D with equal 
weights; the lengths of AC and 
BD being equal. We have, con- 
sequently, between A and B, a 
uniform bending-moment ; while 
on the left of A and on the right 
of B we have a varying bending-moment. The line of bending- 
moments is, in this case, CabD. 

We may, in a similar way, derive curves of bending-moment 
for all cases of loading and supporting beams. 





Fig. 225. 



STRESSES AT DIFFERENT FARTS OF A BEAM. 29 1- 

§ 193. Mode of Procedure for Ascertaining the Stresses 
at Different Parts of a Beam when the Loads and the Di- 
mensions are given, and when no Fibre at the Cross- 
section under Consideration is Strained beyond the 
Elastic Limit. — When the dimensions of a beam, the 
load and its distribution, and the manner of supporting are 
given, and it is desired to find the actual intensity of the stress 
on any particular fibre at any given cross-section, we must pro- 
ceed as follows : - - 

i°. Find the actual bending-moment (M) at that cross-sec- 
tion. 

2°. Find the moment of inertia (/) of the section about its 
neutral axis. 

3 . Observe, that, from what has already been shown, the 
moment of the couple formed by the tensions and compressions 
is af, where a = intensity of stress of a fibre whose distance 
from the neutral axis is unity, and that this moment must equal 
the bending-moment at the section in order to secure equilib- 
rium. Hence we must have 

al= M. 

Moreover, if / denote the (unknown) intensity of the stress 
of the fibre where the stress is desired, and if y denote the 
distance of this fibre from the neutral axis, we shall have 

P P T TUT My 

*=? ••• y = M > •'• P = ^> 

from which equation we can determine/. 

EXAMPLES. 

i. Given a beam 18 feet span, supported at both ends, and loaded, 
uniformly (its own weight included) with 1000 lbs, per foot of length. 
The cross-section is a T, where area of flange = 3 square inches, 
area of web = 4 square inches, height = 10 inches. Find (a) the 



292 APPLIED MECHANICS. 



bending-moment at 3 feet from one end ; (b) the greatest bending- 
moment; (c) the greatest intensity of the tension at each of the 
above sections ; (d) the greatest intensity of the compression at each 
of these sections. 

2. Given an I-beam with equal flanges, area of each flange = 3 
square inches, area of web = 3 square inches, height =10 inches ; the 
beam is 12 feet long, supported at the ends, and loaded uniformly (its 
■own weight included) with a load of 2000 lt>s. per foot of length. Find 
(a) the bending-moment at a section one foot from the end ; (b) the 
greatest bending-moment ; (c) the greatest intensity of the stress at 
each of the above cross-sections. 



§ 191. Mode of Procedure for Ascertaining the Dimen- 
sions of a Beam to bear a Certain Load, or the Load that 
a Beam of Given Dimensions and Material is Capable of 
Bearing. — If we wish to determine the proper dimensions 
of the beam when the load and its distribution, as well as the 
manner of supporting, are given, so that it shall nowhere be 
strained beyond safe limits, or if we wish to determine the 
greatest load consistent with safety when the other quantities 
are given, we must impose the condition that the greatest 
intensity of the tension to which any fibre is subjected shall 
not exceed the safe working-strength for tension of the mate- 
rial of which the beam is made, and the greatest intensity of 
the compression to which any fibre is subjected shall not exceed 
the safe working-strength of the material (or compression. 

Thus, we must in this case first determine where is the 
section of greatest bending-moment (this determination some- 
times involves the use of the Differential Calculus). 

Next we must determine the magnitude of the greatest 
bending-moment, absolutely if the load and length of the beam 
are given (if not, in terms of these quantities), and then equate 
this to the moment of the resisting couple. 

Thus, if M Q is the greatest bending-moment, when the loads 
are such that no fibre is strained beyond the elastic limit, I the 



WORKINGS TRENG Til 293 

moment of inertia of that section where this greatest bending-moment 
acts, and if j t = greatest tensile fibre stress per square inch, j c = 
greatest compressive fibre strength per square inch, y t = distance 
of most stretched fibre from the neutral axis, and y c = distance 

of most compressed fibre from the neutral axis, then will — be 

yt 

the greatest tension per square inch, at" a unit's distance from the 

neutral axis, and — the greatest compression per square inch, at a 

unit's distance from the neutral axis. 

Moreover, in this case, these two ratios are equal, and hence 

M = f± I = f -L 

yt ■ y c 

SAFE OR WORKING-LOAD. 

If // = safe working-strength per square inch for tension, 
/ c '=safe working-strength per square inch for compression, and 
M = greatest safe working bending-moment, then the ratios, 

— and — , are not equal. 

yt yc 

f ' I f f r 

Hence, when — is less than — we have Mq=—L and when 

yt y c y t 

i ' f f / ' 

— is greater than — we have Md =—I. 

yt 6 y c y c 

BREAKING-LOAD AND MODULUS OF RUPTURE. 

If M is the greatest bending-moment when the beam is 
subjected to its breaking-load, the formulae given above do not 
apply, inasmuch as a portion of the fibres are strained beyond 
the elastic limit, and Hooke's law no longer holds, since, after 
the elastic limit is passed, the ratio of stress to strain decreases 
when the stress increases. 

Indeed, the stresses in the different fibres are no longer pro- 



294 APPLIED MECHANICS. 

portional to the distances of those fibres from the neutral axis. 
A graphical representation of the stress at different points of any- 
given section AB would be of the character shown in the figure, 
o the form of the curve CDE varying with the shape 



DJP^' i of the cross-section. 

e b Nevertheless, it is customary to compute the 

breaking-strength of a beam by means of the 

My 
formula j=~zr- J where y is taken as the distance from the neutral 

axis to that outer fibre which gives way first, i.e., to the most 
stretched fibre if the beam breaks by tension, or to the most com- 
pressed fibre, if it breaks by compression. The quantity /, which 
may thus be computed from the formula 

My 

is defined as the Modulus oj Rupture. 

Inasmuch as this formula would give the outside fibre stress, 
if the stress were uniformly varying, it follows that, in the case of 
materials for which the tensile is less than the compressive strength, 
the modulus of rupture is greater than the tensile strength, while 
in that of materials for which the compressive is less than 
the tensile strength the modulus of rupture is greater than the 
compressive strength. 

For experimental work bearing upon this matter, see an 
article by Prof. J. Sondericker, in the Technology Quarterly for 
October, 1888. 

WORKING-STRENGTH. 

The working-strength per square inch of a material for trans- 
verse strength is the greatest stress per square inch to which it 
is safe to subject the most strained fibre of the beam. It is usually 
obtained by dividing the modulus of rupture by some factor of 
safety, as 3 or 4. 



WORKINGS TRENG TH. 2g$ 



§ 192. EXAMPLES. 

i. Given a beam (Fig. 226) supported at both ends, and loaded, 
i°, with w pounds per unit of length uniformly, and 2 , with a single 
load W bX a distance a from the left-hand support: find the position 
of the section of greatest bending-moment, and the value of the greatest 
bending-moment. 
o a b Solution. 



wl W(l — a) 

(1) Left-hand supporti ng- force =' 1 : . 

x 2 / 

FigT 2 6- Right-hand supporting-force = _ _j < L, 

(2) Assume a section at a distance x from the left-hand support 
{this support being the origin), and the bending-moment at that sec- 
tion is, — 

(wl W{1 - a)} wx* 
when x < a, < — H - ? >x ; 

and when x > a, 



W{1 - a) \ wx* ir/ . 
-\ — ->x W(x — a). 



W l w(l - a) 

2 



To find the value of x for the section of greatest bending-moment, 
differentiate each, and put the first differential co-efficient = zero. 
We shall thus have, in the first case, 

wl W(l -a) I W(l - a) 

j -. wx = o, or x = - + 

2 / 2 

and in the second case, 

wl W{1 - a) Tjr I t 

j j wx — yy = o, or x = — f- 

2 » ' 2 

Now, whenever the first is < «, or the second is *> a, we shall have 
in that one the value of x corresponding to the section of greatest 
bending-moment. But if the first is > a, and the second < a, then the 
greatest bending-moment is at the concentrated load. 

These conclusions will be evident on drawing a diagram representing 
the bending-moments graphically, as in Figs. 223 and 224; and the 
greatest bending-moment may then be found by substituting, in the cor- 
responding expression for the bending-moment, the deduced value of or. 



wl 


i 




W{1- 


a) 


W 


wl 




w 



296 APPLIED MECHANICS. 

2. Given an I-beam, 10 feet long, supported at both ends, and 
loaded, at a distance 2 feet to the left of the middle, with 20000 pounds. 
Find the bending-moment at the middle, the greatest bending-moment, 
also the greatest intensity of the tension, and that of the compression at 
each of these sections. 

Given Area of upper flange = 8 sq. in. 

Area of lower flange = 5 sq. in. 

Area of web = 7 sq. in. 

Total depth = 14 in. 

§ 193. Beams of Uniform Strength. — Abeam of uniform 
strength (technically so called) is one in which the dimensions 
of the cross-section are varied in such a manner, that, at each 
cross-section, the greatest intensity of the tension shall be 
the same, and so also the greatest intensity of the com- 
pression. 

Such beams are very rarely used ; and, as the cross-section 
varies at different points, it would be decidedly^ad engineering 
to make them of wood, for it would be necessary to cut the 
wood across the grain, and this would develop a tendency to 
split. 

In making them of iron, also, the saving of iron would gen- 
erally be more than offset by the extra cost of rolling such a 
beam. Nevertheless, we will discuss the form of such beams in 
the case when the section is rectangular. 
• In all cases we have the general equation 

M = tl 

y 

applying at each cross-section, where M = bending-moment 
(section at distance x from origin), / = moment of inertia of 
same section, y = distance from neutral axis to most strained 
fibre, and p = intensity of stress on most strained fibre ; the 
condition for this case being that / is a constant for all values 
of x (i.e., for all positions of the section), while M, f, and j 
are functions of x. 



BEAMS OF UNIFORM STRENGTH. 297 

As we are limiting ourselves to rectangular sections, if we 
let b == breadth and h = depth of rectangle (one or both vary- 
ing with x), we shall have 

M = £bh* 

6 

as the condition for such a beam, with/ a constant for all values 
of x, when the same load remains on the beam. 

We must, therefore, have bh 2 proportional to M. Hence, 
assuming the origin as before, 

i°. Fixed at one end, load at the other, bh 2 =(— ) W(l — x). 

m 

2 . Fixed at one end, uniformly loaded, bh 2 = (- — ) (/- x) 2 . 

\p 2/' 

3 . Supported at ends, loaded at I 2 \p 2 / 

middle. | , . / , , 2 /6 W\, 7 . 

for x > -, M 2 = ( ) (/ — x) . 

\ 2 \p ,2 / 

4 . Supported at ends, uniformly loaded, bh 2 =( )Ux — x 2 ). 

\p 2/' - 

Now, this variation of section may be accomplished in one 
of two ways: 1st, by making h constant, and letting b vary; 
and 2d, by making b constant, and letting h vary. Thus, in 
the first case above mentioned, if h is constant, we have, for the 
plan of the beam, 



*-©«-*>< 



ph 

and if one side be taken parallel to the axis of the beam, this 
will be the equation of the other side ; and, as this is the equa- 
tion of a straight line, the plan will be a triangle. 



2$$ APPLIED MECHANICS. 

If, on the other hand, b be constant, and h vary, we shall 
have, for the vertical longitudinal section of the beam, 



*-©fe* 



and, if one side be taken as a straight line in the direction of 
the axis, the other will be a parabola. 

A similar reasoning will give the plan or elevation respect- 
ively in each case ; and these can be readily plotted from their 
equations. 

CROSS-SECTION OF EQUAL STRENGTH. 

A cross-section of equal strength (technically so called) is 

one so proportioned that the greatest intensity of the tension 

shall bear the same ratio to the breaking tensile strength of the 

material as the greatest intensity of the compression bears to 

the breaking compressive strength of the material. This is 

accomplished, as will be shown directly, by so arranging the 

form and dimensions of the section that the distance of the 

neutral axis from the most stretched fibre shall bear to its 

distance from the most compressed fibre the same ratio that 

the tensile bears to the compressive strength of the material. 

Let f c = breaking-strength per square inch for compression, 

f t — : breaking-strength per square inch for tension, 

y c _= distance of neutral axis from most compressed 

fibre, 

y t = distance of neutral axis from most stretched fibre. 

If p c = actual greatest intensity of compression, and p t = 

actual greatest intensity of tension, then, for a cross-section 

of equal strength, we must have, according to the definition, 

^= <l\ but we have — = — = intensity of stress at a unit's 

Pt ft y c y t 



CROSS-SECTION OF EQUAL STRENGTH. 299 

distance from the neutral axis. Hence, combining these two, 
we obtain 

yt ft 

EXAMPLE. 

Suppose we have/. = 80000 lbs. per square inch, and/ = 20000 
lbs. per square inch. : find the proper proportion between the flange A t 
and the web A 2 of a T-section whose depth is h. 

§ 194. Deflection of Beams. — We have already seen (§ 185), 
that, in the case of a beam which is bent by a transverse load, 
we have 

where (having assumed a certain cross-section whose distance 
from the origin is x) a = the strain of a fibre whose distance 
from the neutral axis is y y and r = radius of curvature of 
the neutral lamina at the section in question. Hence follows the 
equation 

1 __ a m 
— — - '} 

r y 

but from the definition of E y the modulus of elasticity, we shall 

have 

P 
a = -, 

where / = intensity of the stress at a distance y from the 
neutral axis. 

Hence it follows, assuming Hooke's law, that 

1 _ p ____ 1 p 
r Ey E y 

We have already seen, that, disregarding signs, M = - / 

y 



300 APPLIED MECHANICS. 

(making, of course, the two assumptions already spoken of 
when this formula was deduced), where M = bending-moment 
at, and / = moment of inertia of, the section in question ; i.e., 
of that section whose distance from the origin is x. This gives 

*- = , if, denoting tension by the + sign, and taking y 

y J 

positive upwards, we call M positive when it tends to cause 
tension on the lower, and compression on the upper, side; these 
being the conventions in regard to signs which we shall adopt 
in future. Hence, by substitution, we have 

1=-^=' k. (i) 

r Ey ~ EI v ' 

Now, if we assume the axis of x coincident with the neutral 
line of the central longitudinal section of the beam, and the 
axis of v at right angles to this, and v positive upwards, no 
matter where the origin is taken, we shall always have, as is 
shown in the Differential Calculus, 

_d 2 v 
i dx 2 



HS") 



Hence equation (i) becomes 



d 2 v 

~dx 2 M 



(-mi 



I £/ 



(2) 



M and / being functions of x : and, when we can integrate 
this equation, we can obtain v in terms of x, thus having the 
equation of the elastic curve of the neutral line ; and, by com- 
puting the value of v corresponding to any assumed value of x> 
we can obtain the deflection at that point of the beam. 



FORMULA FOR SLOPE AND DEFLECTION. 30 1 

The above equation (2) is, as a rule, too complicated to be 
integrated, except by approximation ; and the approximation 
usually made is the following : — 

Since in a beam not too heavily loaded, the slope, and con- 
sequently the tangent of the slope (or angle the neutral line 
makes with the horizontal at any point), is necessarily small, it 

follows that — is very small, and hence ( — - ) is also very small, 
dx \dxj 

and 1 + ( — ) is nearly equal to unity. Making this substitu- 
tion, we obtain, in place of equation (2), 

d 2 v M , v 

-d^ = Ei'> (3) 

and this is the equation with which we always start in com- 
puting the slope and deflection of a loaded beam, or in finding 
the equation of the elastic line. 

By one integration (suitably determining the arbitrary con- 
stant) we obtain the slope whose tangent is — , and by a second 

dx 

integration we obtain the deflection v at a distance x from the 
origin ; and thus, by substituting any desired value for x, we 
can obtain the deflection at any point. 

§ 195. Ordinary Formulae for Slope and Deflection. — 
We may therefore write, if i is the circular measure of the 

slope at a distance x from the origin, since i = tan i = -— 

- dx 

nearly, 

d 2 v = M 

dx 2 ~ Ef 



j^ 7 



dx . EI 



-ff£* 



302 APPLIED MECHANICS. 

In these equations, of course, E is taken as a constant, M 
must always be expressed in terms of x, and so also must / 
whenever the section varies at different points. When, how- 
ever, the section is uniform, / is constant, and the formulae 
reduce to 

Mdx 2 . 



h$ Mdx > -"hfj 2 



EI 

§ 196. Special Cases i°. Let us take a cantilever loaded 

with a single load at the free end. Assume the origin, as 
before, at the fixed end, and let the beam be one of uniform 
section. We then have M = — W{1 — x), 

. • W Ctt w Wf, x 2 \ , 

To determine c, observe that when x = o, i = o ; 

W/ 7 x*\ 



c — o 



'~5<— 9 <■> 



is the slope at a distance ^r from the origin. 
The deflection at the same point will be 

v = fidx = -^/(* - f)^ = -g(¥ - f) + " 

but when ;r = o, v = o .*. c — o .*. the deflection at 
a distance ;? from the origin will be 

"=-1/(7-6-)' (2) 

The equations (1) and (2) give us the means of finding the 
slope and deflection at any point of the beam. ' 

To find the greatest slope and deflection, we have that both 
expressions are greatest when x = /. Hence, if i and v rep- 
resent the greatest slope and deflection respectively, 

Wl 2 Wl* 

/o " 7EI' Vo ~ lEf 



SPECIAL CASES. 3°3 



2°. Next take the case of a beam supported at both ends 
and loaded uniformly, the load per unit of length being w. 
Assume the origin at the left-hand end ; then 



and W ' = wl 
w /lx 2 x^ 



wl wx z 

I = — -x — 

2 2 


= 


w 
-Ux - 

2 V 


X 2 ) 




W I 


{lx 


' — x 2 )dx 


= 


w 
lEI 



/lx 2 x*\ 

(t-t) + ^ 



/ 

To determine c, we have that when x = -, then i = o: 
w //3 /3\ ze//3 

.*• O = — F- r ( ) + € .'. C — 



2EI \ 8 24/ 2\EI 

w /lx 2 x 3 \ wl* w 



1 ~~ 2EI 



/lx* X*\ Wl* w 

(t - 7) - ^m = ^m^ ~ ^ ~ /3 > « 



24^/ 
But when .r = o, v = o ; 



{2CX* _ #4 -_ /3#) -(- tf# 



C = O 






For the greatest slope, we have x = o, or ;r = // 

W/3 -Jf7 2 



2a = 



24^/ 2^EI 



For the greatest deflection, x = - ; 

2 



— w 5/ 4 — 5W/ 4 —$Wfi 



24E/ 16 384^/ " 384^/ 



304 APPLIED MECHANICS. 

3°. Take the case of a beam supported at both ends, and 
loaded at the middle with a load W. 

Assume, as before, the origin at the left-hand support. 
Then we shall have 

W I W I 

M = —x, x < -, and M = — (/ — x) when x> — 

2 2 2 V y 2 



Therefore, for the slope up to the middle, we have 

W x 2 
2EI 2 



w r w x 2 



l 

When x = — , then 1 ' = o ; 
2 

wr 



- - 16EI'. 

•' % - A EI\ X J> (I) 

and 

^ /V . /2N ^ ^^ s /2 ^ . 

But when x =. o, v = o ; 



c = o. 



). (2) 



4M3 4 

The slope is greatest when x — o\ 

. _ - ivr 
- *°~ 16EI' 

1 

The deflection is greatest when x = -; 

_ - ET/ 3 
" * *° 48^/ ' 

4 . In the following table / denotes the moment of inertia 
of the largest section : 



SPECIAL CASES. 



305 



Uniform Cross-Section. 



Greatest Slope. 



Greatest 
Deflection. 



Fixed at one end, loaded at the other 

Fixed at one end, loaded uniformly . . 

Supported at ends, load at middle . . . 

Supported at ends, uniformly loaded . . 



1 Wl 2 
*~EJ 
1 Wl 2 
*EI 
1 Wl* 
16 EI 
1 Wl 2 
24 EI 



Uniform Strength and Uniform Depth, 
Rectangular Section. 



1 wi* 

3 EI 

1 Wl* 
8 EI 
1 Wl* 

48 EI 
5 Wl* 
384 EI 



Fixed at one end, load at the other . . 
Fixed at one end, uniformly loaded . . 
Supported at both ends, load at middle . 
Supported at both ends, uniformly loaded, 



Wl 2 


1 Wl* 


EI 


2 EI 


1 Wl 2 
2 ~EI 


1 Wl* 
4 EI 


1 Wl 2 
&EI 


1 Wl* 
32 EI 


1 Wl 2 
16 EI 


1 Wl* 
64 EI 



Uniform Strength and Uniform Breadth, 
Rectangular Section. 



Fixed at one end, loaded at the other, 
% Supported at both ends, load at middle, 



wr 

2 ~EI 

1 wr 



4 EI 

I Wl* 

Supported at both ends, uniformly loaded 0.098 —=— 



2ivr 

3 EI 

1 wr 

24 EI 

wr 



0.018 



EI 



306 APPLIED MECHANICS. 

§197. Deflection with Uniform Bending-Moment. — If 

the bending-moment is uniform, then M is constant ; and, if / 
is also constant, we have 



M r Mx 

~ ~EI J dx ~ ~EI 



+ c; 



I 
but when x = -, then i = o ; 

Ml 



M (x 1 lx\ 






the constant disappearing because v = o when j*r = o. 

Hence, for a beam where the bending-moment is uniform, 

we have 

Mi l\ M/x* /x\ 

> = £7\ x -2)> v = El\^--2)> 

and for greatest slope and deflection, we have 



■Ml M/l* l 2 \ 1 Ml 



la = -FT, V Q = 






7)- 



72 



EI' "°~£I\8 4/ ~ 8^/ 

§ 198. Resilience of a Beam. — The resilience of a beam 
is the mechanical work performed in deflecting it to the amount 
it would deflect under its greatest allowable gradually applied 
load. In the case of a concentrated load, if W is the greatest 
allowable gradually applied load, and v l the corresponding 
deflection at the point of application of the load, then will the 

n • 1 W 

mean value of the load that produces this deflection be — > 

W 
and the resilience of the beam will be ~zv v 



SLOPE AND DEFLECTION OF A BEAM. 307 

§ 199. Slope and Deflection of a Beam with a Con- 
centrated Load not at the < > 

Middle. — Take, as the next ° b. % 

case, a beam (Fig. 228). Let 
the load at A be W, and dis- 
tance OA = a, and let a > -. 

2 

W{1 - a) 
x < a M — — ^ '- x, 

x>a M=^(/-x), 



, , 


< 


a > 




■■ ■ /v 




Fig. 228. 





x<a i = m mr l $ xdx = l ^iEr l * + e ' 



- I xdx — 

When x = o, i = i Q = undetermined slope at O; 

. W{1 - a) % , . , x 



2lEI 



and 



When ^ = o, z> = o ; 

_ W{1 - a) 



61EI 



x* + t'oX, (2) 



To determine ^, observe that when x = #, this value of # 
and that deduced from (1) must be identical. 



Wal, a 2 \ , tfTZ-aW , . «^ 2 , . 

mr-T) + '- 2IE1 +'° •*• '--s? + <- 



30S APPLIED MECHANICS. 



Wa . x 2 \ 
,\ I = - [ Ix ] 

IE1\ 2 ) 



Wa 2 



or 



i = -^§} ( < 2lx ~ & ~ la ) + '« (3) 

and 

v = ^- I (2/x — x 2 — /#)<&; + 4 / ^ 
2ZEIJ J 

Wa , J7 _ x* / \ , • 
= -yw-ilx 2 lax) + tax + c. 

2l£Ll 3 

To determine c, observe that when x = a, this value of v 
and (2) must be identical ; 



Wa / a\ . W(/ - a) 






2/^/\ 3 

^ „ , A 4 x #7# 3 Wa> 

/. v = -j^Ulx 2 - a* - 3/** + ^ 2 ) + 4*. (4) 

To determine 4> we have that when ;r = /, z> = o ; 
0fc 



61EI 



(2/3 - 30/3 + la 2 ) + 4/ 



... / o = _i^_( 3 ^/ 2 _ 2 /3 _ la 2 ) = ^-(xal - 2/ 2 - a*). 
6l 2 EI K6 6/E/ Kd J 

Substituting this value of i Q in the equations (1), (2), (3), and 
(4), we obtain for 

IV(1- a) 2 Wa . 7 72 2 . 

(1) 1 = — i L x * _f ( W _ 2 / 2 — a 2 ), 

V ' 2IEI 61EI 6 ; 

(2) z> = ' X* H (30/ — 2/ 2 — tf 2 )*, 

■ ; 61EI 6lEl y6 } 



SLOPE AND DEFLECTION OF A BEAM. 309 



(3) * = Jl (2/ * -*-*> + Mi^ a/ - 2/2 - a2) > 

(4) v = M/ (s/x2 ~ xz " zlax + /a2) + M/ {3a/ " 2/2 ~ * 2) *- 

To find the greatest deflection, differentiate (2), and place 
the first differential co-efficient equal to zero : or, which is the 
same thing, place i = o in (1), and find the value of x ; then 
substitute this value in (2), and we shall have the greatest 
deflection. 

We thus obtain 

(/-*)** = ^( 3 a/ -2/.-*) ... * = *( ** -S't+t S 
3 3\ l — ' / 

or 



x* = a -(2l-a) /. x ^ 2al - a " \ 
3 V3 



and the greatest deflection becomes 



Wa{l - a) (2/ - a) VW - a* 



§ 200. EXAMPLES. 

1. In example 1, p. 294, find the greatest deflection of the beam 
when it is loaded with J of its breaking-load, assuming E = 1200000. 

2. In the same case, find what load will cause it to deflect ^q- of its 
span. 

3. What will be the stress at the most strained fibre when this occurs* 

4. In example 3, p. 294, find the load the beam will bear without 
deflecting more than ^J^ ot its span, assuming E — 24000000. 

5. Find the stress at the most strained fibre when this occurs. 

6. In example 6, p. 295, find the greatest deflection under a load 
J- the breaking-load. 



3io 



APPLIED MECHANICS. 



§ 201. Deflection and Slope under Working-Load. — If 
we take the four cases of deflection given in the first part of 
the table on p. 305, and calling/ the working strength of the 
material, and y the distance of the most strained fibre from 
the neutral axis, and if we make the applied load the working- 
load, we shall have respectively — 



Wl = 


fl 

y 


Wl 

2 


fl 

y 


Wl 
4 


fl 

y 


Wl 


fl 

y 



4. -0- = - 



w= 


fl 

' iy' 


w= 


2/1 

' iy' 


w= 


■ ly' 


w= 


S/I 

iy' 



And the values of slope and deflection will become respectively, 





Slope. 


Deflection. 




Slope. 


Deflection. 


1°. 

2°. 


*iy 

zJ Ey 


V-gy 

V- 

iJ Ey 


3°- 
4°. 


id. 

± J Ey 
; / 


r *fiy 

1 2 

_5_ /__ 

** J Ey 



From these values, and those given on p. 305, we derive the 
following two propositions : — 

i°. If we have a series of beams differing only in length; 
and we apply the same load in the same manner to each, their 
greatest slopes will vary as the squares of their lengths, and 
their greatest deflections as the cubes of their lengths. 



SLOPE AND DEFLECTION OF RECTANGULAR BEAMS. 3II 

2°. If, however, we load the same beams, not with the same 
load, but each one with its working-load, as determined by 
allowing a given greatest fibre stress, then will their greatest 
slopes vary as the lengths, and their greatest deflections as the 
squares of their lengths. 

§ 202. Slope and Deflection of Rectangular Beams 

bh> h 

If the beams are rectangular, so that / = — and y = -, the 

values of slope and deflection above referred to become further 
simplified, and we have the following tables : — 





Given Load IV. 


Working-Load. 
Greatest Fibre Stress =/. 




Slope. 


Deflection. 


Slope. 


Deflection. 


1°. 


6W1 2 
Ebh* 


Ebh* 


ft 

Eh 


3 Eh 


2° 


2WI* 


3 Wl* 
2 Ebh* 


1A 

ZEh 


ifl 2 




Ebfc 


*Eh 


3°- 


3 ^ 2 
iEbfc 


1 m* 

4£bfr 


1/ 
2 Eh 


ifl 2 
6 Eh 


4°- 


1 WI* 

2 Ebh* 


5 #73 
32 Ebh* 


2JI 

3 Eh 


±2L 

24 Eh 



So that, in the case of rectangular beams similarly loaded and 
supported, we may say that — 

Under a given load W, the slopes vary as the squares of 
the lengths, and inversely as the breadths and the cubes of the 
depths ; while the deflections vary as the cubes of the lengths, 
and inversely as the breadths and the cubes of the depths. 



312 



APPLIED MECHANICS. 



On the other hand, under their working-loads, the slopes vary 
directly as the lengths, and inversely as the depths ; while the 
deflections vary as the squares of the lengths, and inversely as 
the depths. 

§ 203. Beams Fixed at the Ends. — The only cases which 
we shall discuss here are the two following ; viz., — 
i°. Uniform section loaded at the middle. 
2°. Uniform section, load uniformly distributed. 
Case I. — Uniform Section loaded at the Middle. — The 
fixing at the ends may be effected by building the beam for 

some distance into the wall, as 
shown in Fig. 229. The same 
result, as far as the effect on 
the beam is concerned, might 
be effected as follows : Hav- 
ing merely supported it, and 



w 



Fig. 229. 



placed upon it the loads it has to bear, load the ends outside 
of the supports just enough to make the tangents at the sup- 
ports horizontal. 

These loads on the ends would, if the other load was re- 
moved, cause the beam to be convex upwards : and, moreover, 
the bending-moment due to this load would be of the same 
amount at all points between the supports ; i.e., a uniform 
bending-moment. Moreover, since the effect of the central 
load and the loads on the ends is to make the tangents over 
the supports horizontal, it follows that the upward slope at the 
support due to the uniform bending-moment above described- 
must be just equal in amount to the downward slope due to the 
load at the middle, which occurs when the beam is only sup- 
ported. 

Hence the proper method of proceeding is as follows : — 
i°. Calculate the slope at the support as though the beam 
were supported, and not fixed, at the ends ; and we shall have, 
if we represent this slope by i„ the equation 



BEAMS FIXED AT THE ENDS. 313 



«BI (I) 



2°. Determine the uniform bending-moment which would 
produce this slope. 

To do this, we have, if we represent this uniform bending- 
moment by M iy that the slope which it would produce would be 

(2) 



2El y 
and, since this is equal to t„ we shall have the equation 

MJ Wl* 



2 EI 16EI 



= (3) 



... M x = --. (4) 

This is the actual bending-moment at either fixed end ; and the 
bending-moment at any special section at a distance x from 
the origin will be 

M+M v 

where M is the bending-moment we should have at that sec- 
tion if the beam were merely supported, and not fixed. Hence, 
when it is fixed at the ends, we shall have, for the bending- 
moment at a distance x from O, where O is at the left-hand 
support, 

W W 
M=^ X -^l. (5) 

When x = -, we obtain, as bending-moment at the middle, 
2 

M = — ; (6) 

and, since M x = — M oi it follows that the greatest bending- 
moment is 

W2 

8'" 



3 T 4 APPLIED MECHANICS. 

this being the magnitude of the bending-moment at the middle 
and also at the support. 

POINTS OF INFLECTION. 

The value of M becomes zero when 

x = - and when x = — ; 
4 4 

hence it follows that at these points the beam is not bent, and 
that we thus have two points of inflection half-way between the 
middle and the supports. 

SLOPE AND DEFLECTION UNDER A GIVEN LOAD. 

We shall have, as before, 



CM , W& Wlx , 




and since, when x = o, i = o, 




.*. c = o 




. dv W , _ 7V 

dx 8£I K ' 


(7) 


W (2X* lx 2 \ 


(8) 



the constant vanishing because v = o when x = o. The slope 

becomes greatest when x = -, and the deflection when x = -. 

4 2 

Hence for greatest slope and deflection, we have 

10 = "^/' (9) 

»o = «>. (10) 

192-fiy 



BEAMS FIXED AT THE ENDS. 315 



SLOPE AND DEFLECTION UNDER THE WORKING-LOAD. 

If f represent the working-strength of the material per 
square inch, and if W represent the centre working-load, we 
shall have 

8 ~ y 
, W^Il (xx) 



• *o 



'—-YEy < I2 > V ^~T,Wy (I3) 



Case II. — Uniform Section, Load uniformly Distributed. — 
Pursuing a method entirely similar to that adopted in the former 
case, we have — 

1°. Slope at end, on the supposition of supported ends, is 

Wl* , x 

* = ~^EI (I) 

2°. Slope at end under uniform bending-moment M x is 

_MJ 
2EI 

Hence, since their sum equals zero, 



w 



M ' = "IP <5) 

which is the bending-moment over either support. 
The bending-moment at distance x from one end is 

W W ,7 , Wl , . 

J/= ~ (/* — * 2 ) . (4) 

2/ 12 

This is greatest when x = o, and is then . Hence great- 
est bending-moment is, in magnitude, 

H . (5) 

12 



3l6 APPLIED MECHANICS. 



POINTS OF INFLECTION. 

M becomes zero when x = - ± — =. (6) 

2 2 v/3 



Hence the two points of inflection are situated at a distance 
/ 

2^ 



— = on either side of the middle. 



'-■•/ 



SLOPE AND DEFLECTION. 

M W 



the constant vanishing because z = o when x = o. 

W \, . x* 



12IEI 



K- ?'-?}• < 8 > 



the constant vanishing because ^ = o when ;r = o. Hence for 
greatest slope and deflection we have, i is greatest when x = 

~( 1 zb -7=1, and z; is greatest when x = - ; 

2\ v 3 ; 2 

12sj3EI 
*° - ~^Ef (I °> 



SLOPE AND DEFLECTION UNDER WORKING-LOAD. 

For working-load we have 

m fi 

12 y 



(") 



W = *-¥ ■ <„) 



6\flEy 



(13) 



% = —4". (14) 



BENDING-MOMENT AND SHEARING-FORCE. 317 



EXAMPLES. 

i. Given a 4-inch by 12-inch yellow-pine beam, span 20 feet, fixed 
at the ends ; find its safe centre load, its safe uniformly distributed load, 
and its deflection under each load. Assume a modulus of rupture 5000 
lbs. per square inch, and factor of safety 4. Modulus of elasticity, 
1200000. 

2. Find the depth necessary that a 4-inch wide yellow-pine beam, 20 
feet span, fixed at the ends, may not deflect more than one four-hun- 
dredth of the span under a load of 5000 lbs. centre load. 

§ 204. Variation of Bending-Moment with Shearing- 
Force. — If in any loaded beam whatever, M represent the 
b ending-mom enty and F the shearing-force at a distance x from 
the origin, then will 

*-% (,) 

Proof (a). — In the case of a cantilever (Fig. 230), assume 
the origin at the fixed end ; then, if M 
represent the bending-moment at a 
distance x from the origin, and M -\- AM 
that at a distance x -f- Ax from the 
origin, we shall have the following 
equations: — 

x = l 

M = -2 W(a- x), 

X = X 

x = l 

M+ AM = — 2 W{a— x — Ax) nearly. 

X = X 

a being the co-ordinate of the point of application of W t 

X = i 

. AM = Ax% JF nearly 

X — X 

... ^=rv, 

Ax *=~ 



3 I 8 APPLIED MECHANICS. 



and, if we pass to the limit, and observe that 

x = l 

X = X 

we shall obtain 



* 



(#) In the case of a beam supported at the ends (Fig. 231), 
._«.-* assume the origin at the left-hand 



.jj 7s, end, and let the left-hand support- 

ing-force be S ; then, if a represent 
fig. 231. the distance from the origin to the 

point of application of W, we shall have the equations 



M=Sx-$ W(x - a), 

x — o 
x = x 

M + AM = S(x -f Ax) - 2 W(x - a + A*) nearly. 

x = o 

Hence, by subtraction, 

X = X 

AM = S . A* — 2 WAx nearly 



ai/ ■* = •* 

— = £-2 ^nearly; 

Ax 



x = o 



and, if we pass to the limit, and observe that 

X = X 

F = S - 2 W, 



we shall obtain 



as before. 



£-'• <*> 



\ 

c 


« b 

\ 1 


•J 

'a 



LONGITUDINAL SHEARING OF BEAMS. 319 

§ 205. Longitudinal Shearing of Beams. — The resistance 
of a beam to longitudinal shearing sometimes becomes a mat- 
ter of importance, especially in timber, where the resistance to 
shearing along the grain is very small. We will therefore pro- 
ceed to ascertain how to compute the intensity of the longi- 
tudinal shear at any point of the beam, under any given load ; 
as this should not be allowed to exceed a certain safe limit, to 
be determined experimentally. Assume a 
section A C (Fig. 232) at a distance x from 
the origin, and let the bending-moment at 
that section be M. Let the section BD be 
at a distance x -\- Ax from the origin, and _ 
let the bending-moment at that section be fig. 232. 

M + AM. 

Let y be the distance of the outside fibre from the neutral 
axis; and let ca = y t be the distance of a, the point at which 
the shearing-force is required, from the neutral axis. 

Consider the forces acting on the portion ABba, and we 
shall have — 

My Q 
i°. Intensity of direct stress at A = -j-. 

2 . Intensity of direct stress at a unit's distance from neu- 

, • M 
tral axis = -=-. 

My 
3 . Intensity of direct stress at e, where ce = y, is -j-. 

„,.,.. , ,. , . (M+ AM)y 
So, likewise, intensity of direct stress at / is j . 

Therefore, if z represent the width of the beam at the point 

e, we shall have — 

^ ■ r M fyo 

Total stress on face Aa = -j J yzdy, 



yi 



Total stress on face Bb 



M+ AM 



f yzdy; 

«/Vi 



320 APPLIED MECHANICS. 

.: Difference = — — / yzdy: 

and this is the total horizontal force tending to slide the piece 
AabB on the face ab. 

Area of face ab, if z t is its width, is 

ZjAx; 

therefore intensity of shear at a is approximately 

AM Cyo 

~rj n yzd y 

z z Ax ' 
or exactly (by passing to the limit) 

fdM\ 

irL yzdy - 

And, observing that F = -j- f this intensity reduces to 



F Cyo 

-A yzdy. (i) 



F 

z 



We may reduce this expression to another form by observ- 
ing, that, if y 2 represent the distance from c to the centre of 
gravity of area Aa, and A represent its area, we have 

pyo 

J yzdy = y 2 A ; 

therefore intensity of shear (at distance^ from neutral axis) at 

point a = 

This may be expressed as follows : — 



LONGITUDINAL SHEARING OF BEAMS. 32 1 

Divide the shearing-force at the section of the beam under 
consideration, by the product of the moment of inertia of the 
section and its width at the point where the intensity of the 
shearing force is desired, and multiply the quotient by the statical 
moment of the portion of the cross-section between the point in 
question and the outer fibre ; this moment being taken about the 
neutral axis. The result is the required intensity of shear. 

The last factor is evidently greatest at the neutral axis ; 
hence the intensity of the shearing-force is greatest at the 
neutral axis. 

LONGITUDINAL SHEARING OF RECTANGULAR BEAMS. 

For rectangular beams, we have 

Hence formula (2) becomes 

12F 

For the intensity at the neutral axis, we shall have, therefore,. 
\2F (h bh\ 3 F 

since for the neutral axis we have 



i2F/h bh\ _$F 

b*h> \ 4 2 / 2 bN ^ 



h , bh 

y 2 = - and A = — . 
4 2 



EXAMPLES. 



1. What is the intensity of the tendency to shear at the neutral axis 
of a rectangular 4-inch by 12-inch beam, of 14 feet span, loaded at the 
middle with 5000 (bs. 



322 



APPLIED MECHANICS. 



2. What is that of the same beam at the neutral axis of the cross- 
section at the support, when the beam has a uniformly distributed load 
of T2000 lbs. 

3. What is that of a 9-inch by 14-inch beam, 20 feet span, loaded 
with 15000 lbs. at the middle. 



§ 206. Strength of Hooks. — The following is the method 
to be pursued in determining the stresses in a 
hook due to a given load ; or, vice versa, the 
proper dimensions to use for a given load. 

Suppose (Fig. 233) a load hung at E ; the 
load being P y and the distances 
AB = n; 

QF = y v 

O being the centre of gravity of this 
section, conceive two equal and opposite 
forces, each equal and parallel to P, acting 

at a 

Let A = area of section, and let / = its 
moment of inertia about CD (BCDF represents the section 
revolved into the plane of the paper) ; then — 

1 °. The downward force at O causes a uniformly distributed 
stress over the section, whose intensity is 




Eig. 233. 



2 . The downward force at E and the upward force at O 

constitute a couple, whose moment is 

Pln+yb; 

and this is resisted, just as the ben ding-moment in a beam, by 
a uniformly varying stress, producing tension on the left, and 
compression on the right, of CD. 



COLUMNS. 323 



If we call p^ the greatest intensity of the tension due to 
this bending-moment, viz., that at B, we have 

and if p % denote the greatest intensity of the compression due 
to the bending moment, viz., that at F, we have 

P% — 7 » 

therefore the actual greatest intensity of the tension is 
/( = /i+A = J + ^±2^ ( 

and this must be kept within the working strength if the load 
is to be a safe one ; and so also the actual greatest intensity" 
of the compression, viz., that at F, is, when/, > p lf 

A A A- 7 ^> 

which must be4cept within the working strength for compression. 

§ 207. Strength of Columns. — The formulae most commonly 
employed for the breaking-strength of columns subjected to a load 
whose resultant acts along the axis have been, until recently, 
the Gordon formulae with Rankine's modifications, the so-called 
Euler formulae, and the avowedly empirical formulae of Hodg- 
kinson. These formulae do not give results which agree with 
those obtained from tests made upon such full-size columns as 
are used in practice. 

The deductions of the first two are not logical, certain assump- 
tions being made which are not borne out by the facts. 

When a column is subjected to a load which strains any fibre 
beyond the elastic limit, the stresses are not proportional to the 
strains, and hence there can be no rational formula for the break- 
ing-load. 

Hence, all formulae for the breaking-load are, of necessity 



3 2 4 APPLIED MECHANICS. 

empirical, and depend for their accuracy upon their agreement 
with the results of experiments upon the breaking-strength of 
such full-size columns as are used in practice. 

Nevertheless, the ordinary so-called deductions of the Gor- 
don, and of the so-called Euler formulae will be given first. 

§ 208. Gordon's Formulae for Columns.— (a) Column fixed in 
c Direction at Both Ends. — Let CAD be the central axis of the 

column, P the breaking-load, and v the greatest deflection, AB. 

Conceive at A two equal and opposite forces, each equal to P; 

then — 
b i°. The downward force at A causes a uniformly distributed 

stress over the section, of intensity, 

2 . The downward force at C and the upward force at A 
Fig. 234. constitute a bending couple whose moment is 

M=Pv. 

If p 2 =the greatest intensity of the compression due to this bending, 

where y= distance from the neutral axis to the most strained fibre of 
the section at A . Then will the greatest intensity of stress at A be 

P , Pvy 

and, since P is the breaking-load, p must be equal to the breaking- 
strength for compression per square inch=/'. 



'-'(■+?> . <■> 



P 

where p= smallest radius of gyration of section at A. 

Thus far the reasoning appears sound; but in the next step it is 
assumed that 

_ ll l 

v = 7y 



GORDON'S FORMULAE FOR COLUMNS. 



325 



where c is a constant to be determined by experiment. Hence, sub-' 
stituting this, and solving for P, 

/A 



P = 






(2) 



which is the formula for a column fixed in direction at both ends. 

(b) Column hinged at the Ends. — It is assumed that the points of 
inflection are half-way between the middle and the ends, and 
hence that, by taking the middle half, we have the case of bending 
of a column hinged at the ends (Fig. 235). Hence, to obtain 
the formula suitable for this case, substitute, in (2), 2/ for I, and 
we obtain 

Fig. 235. 

M 



P = 



1-+ 






(3) 



(c) Column fixed at One End and hinged at the Other (Fig. 236). — 

" In this case we should, in accordance with these assumptions, 

take f of the column fixed in direction at both ends; hence, to 

obtain the formula for this case, substitute, in (2), f/ for /, and 

we thus obtain 



1 + 



16/ 2 

gcp 2 

Fig. 236. 

Rankine gives, for values of / and 
Hodgkinson's experiments: 



(4) 



the following, based upon 



Wrought-iron 
Cast-iron 
Dry timber . 



f 

(lbs. per sq. in.). 



36000 

80000 

7200 



36000 
6400 
3000 



326 



APPLIED MECHANICS. 



§ 2oSa. So-called Euler Formulae for the Strength of 
Columns. 

(a) Column fixed in Direction at One End only, which bends, as 
shown in the Figure. 

i°. Calculate the breaking-load on the assumption that the column 
will give way by direct compression. This will be 

**~/A, (i) 

where f= crushing-strength per square inch, and A = area of cross- 
section in square inches. 

2°. Calculate the load that would break the column if it were to 
give way by bending, by means of the following formula : 

K = (^M (») 

where E= modulus of elasticity of the material, /= smallest moment 
of inertia of the cross-section, and /= length of column. 

Then will the actual breaking-strength, according to Euler, be the 
smaller of these two results. 

To deduce the latter formula, assume the origin at the 
upper end, and take x vertical and y horizontal. 

Let p= radius of curvature at point (x, y), and let 
M = bending-moment at the same point. 

Then we have, with compression plus and tension 
minus, 



Fig. 237. 



Eut 





i 


M 


Py 




P 


EI" 


EI' 




i 
p ~ 


d*y 
~"dx^ 


nearly 


d*v 


P 






~ 




y, 




dx 2 


El 





(3) 



J dx dx 2 EIJ J dx 

(dyV _P_ 

\dxj " EI' 



and, since for y = a, 



r f + c; 
dy 



dx 



= o, 



C = EI*' 



EULER FORMULAE FOR STRENGTH OF COLUMNS. ^ 2 7 

*y J p , 

\la 2 - y 2 V EI 

.-. sin" 1 ^ = U — x + e. 
a t EI 

And since, when x=o, ?=©> .'. c=o, we have 

When j=a, #=/; hence, substituting in (5), and solving for P y 

(6) Column hinged at Both Ends (Fig. 235). 
i°. For the crushing-load, 

2 . For the breaking-load by bending, put I/2 for / in (6) ; hence 



'-.(J)". (7) 



.(c) Column fixed in Direction at One End, and hinged at the other 
(Fig. 236). 

i°. For the crushing-load, 

2 . For the breaking-load by bending, put //3 for /in (6) ; hence 
P 2 = 



-Kt) W - < 8 > 



{d) Column fixed in Direction at Both Ends (Fig. 234). 
i°. For the crushing-load, 

P,=fA. 

2 . For the breaking-load by bending, 

*- (y) EI, ( 9 ) 

;his being obtained from (2) or (6) by substituting I/4 for /. 



328 APPLIED MECHANICS. 

(e) In order to ascertain the length wnere incipient flexure occurs, 
according to this theory we should place the two results equal to each 
other, and from, the resulting equation determine /. We should thus 
obtain, for the three cases respectively, 

/EI 
(a) /= -V j£> (10) 

/Wt 
(/?) l =^jA' (II) 

(r) /=OT ^f- (I2) 

Hence all columns whose length is less than that given in these 
formulae will, according to Euler, give way by direct crushing; and 
those of greater length, by bending only. 

§ 209. Hodgkinson's Rules for the Strength of Columns. 
— Eaton Hodgkinson made a large number of tests of small columns, 
especially of cast-iron, and deduced from these tests certain empirical 
formulae. The strength of pillars of the ordinary sizes used in practice 
has been computed by means of Hodgkinson's formulae, and tabulated 
by Mr. James B. Francis; and we find in his book the following rules 
for the strength of solid cylindrical pillars of cast-iron, with the ends 
flat, i.e., "finished in planes perpendicular to the axis, the weight 
being uniformly distributed on these planes": 

For pillars whose length exceeds thirty times their diameter, 

^=9931877^. (O 

where D= diameter in inches, /= length in feet, W= breaking- weight 
in lbs. 

If, on the other hand, the length docs not exceed thirty times the 
<liameter, he gives, for the breaking-weight, the following formula: 

W = Wc , (2) 

where W= breaking- weight that would be derived from the preceding 
formula, W= actual breaking- weight, 



fes. 



= logSoif^lX ($\ 



BREAKING-LOAD OF FULL-SIZE COLUMNS 329 

For hollow cast-iron pillars, if D= external diameter in inches, d= 
internal diameter in inches, we should have, in place of (1), 

^=99318 — — — , (4) 

and in place of (3), 

c — 109801 — i '-. (5) 

4 

For very long wrought-iron pillars, Hodgkinson found the strength 
to be 1.745 times that of a cast-iron pillar of the same dimensions; but, 
for very short pillars, he found the strength of the wrougi it-iron pillar 
very much less than that of the cast-iron one of the same dimensions. 
With a length of 30 diameters and flat ends, the wrought-iron exceeded 
the cast-iron by about 10 per cent. 

§ 210. Breaking-load of Full-size Columns.— The tests 
made upon full-size columns are not as many as would be desir- 
able. The details will be given in Chapter VII, but a few of the 
empirical formulas which represent their results will- be given here. 

If P = breaking-load, A = area of smallest section, / = length 
of column, p = least radius of gyration of section, and j c = crushing- 
strength of the material per unit of area, it will be found that for 
values of l/p less than a certain amount, the column remains 
straight, and the breaking-load may be computed by means of 
the formula P = ) C A . ' 

For greater values of l/p, the breaking-load is smaller than that 
given by this formula, and may be computed by meani; of the 
formula P = )A, 

by using for / a value smaller than / c , this value varying with the 
value of l/p, and being determined empirically from the results of 
tests of full-size columns. 

(a) In the case of cast-iron columns no tests have been made 
of full-size columns of the second class, while those made upon 
the first class indicate that the value of f c suitable for use in 
practice is from 25,000 to 30,000 lbs. per square inch. 

(b) In the case of wrought-iron columns, the tests of the first 
class indicate that the value of f c suitable for use in practice is 
from 30,000 to 35,000 lbs. per square inch. 



35^ APPLIED MECHANICS 

(c) In the case of wrought-iron columns of the second class, 
the formula of Mr. C. L. Strobel for bridge columns with either 
flat or pin ends, when l/p> go, is 

P a * 

— - = 40000 — 1 2 <— . 

* P 

On the other hand, those recommended by Prof. J. Sonde- 
licker, of which the first was devised by Mr. Theodore Cooper, 
are as follows: 

(a) For Phcenix columns with flat ends l/p > 80, 
P _ 36000 
A~ if {1/p- &>)» • 

l8000 

(/?) For lattice columns with pin-ends and l/p> 60, 
P 34000 

A = ~(l/p-6oF 

12000 

(y) For solid web, square, or box columns with flat ends, and 
l/p>8o, 

P 33000 

3" = I , (1/P-&0) 2 ' 

s 10000 

(d) For solid web, square, or box columns with pin-ends, and 
ljp> 60, 

P _ 31000 

T~ j | (//p-6o)» ' 

6000 

The number of tests that have been made upon full-size steel 
columns is very small, hence no formulae will be given here, but 
the subject will be discussed in Chapter VII. The number of 
tests that have been made upon full-size timber columns is con- 
siderable, but this subject will also be discussed in Chapter VII. 

§ 211. Columns subjected to Loads which do not Strain 
any Fibre beyond the Elastic Limit. — Under this head will 
be discussed, first, the mode of determining the greatest fibre 



THEORY OF COLUMNS. 331 

stress in a straight column subjected to an eccentric load, and, 
secondly, the general theory of columns. 

(a) Straight column, under eccentric load. — Let O' be the 
centre of gravity of the lower section, and let A'O' = x , where 

A' is the point of application of the resultant of the 
I eccentric load. Conceive two equal and opposite 

ijf^p forces at O', each equal and parallel to P. Then we 
have: 

i°. Downward force along 00' causes uniform 

,. . p 

stress of intensity p\—~r* 

2°. The other two form a couple whose moment 

is Px , and the greatest intensity of the stress due to 

(Pxq) a 
this couple is p 2 = r — > where a = 0'B'. Hence, 

Fig. 238. J 

the greatest intensity of the stress is 
P Px a 

and this should be kept within the limits of the working-strength. 

(b) Theory of columns. — The theory of columns is that of 
the Inflectional Elastica, and is explained in several treatises, 
among which is that of A. E. H. Love on the Theory of Elasticity. 
It is as follows: 

Let the curve OP be an elastic line, on which O is a point of 
inflection.' It follows that there is no bending-moment at this 
point, and hence we may assume 
that at O a single force R acts. Take 
the origin at O, and axis of X along the 
line of action of the force R. Let 
Ei = modulus of elasticity of the 
material, 7 = moment of inertia of 
section about an axis through its centre of gravity, and perpen- 




33 2 APPLIED MECHANICS. 

dicular to the plane of the curve, = angle between OX and the 
tangent at any point P whose coordinates are x and y, a = value 
of (j) at point O, r= radius of curvature of the curve at P, s = 
length of arc OP, 1 = length of one bay, i.e., measured from O to 

the next point of inflection, a=y, A= area of section, p =a/t> 
R 



Then we have for any such elastic line, when compressions are 
plus and tensions minus, 

i_M 
P ~EV 

I d(b 

Moreover, since — = —r~ and M = Ry> we have, for a column 

d(f> R 

of the same cross-section throughout its length, —= —-=rjy, 

where the quantity -^-j is a constant. 

By differentiation we obtain 

d 2 4> R dy R . , 

d^ = ~EjTs = -Ej Sm(l) ' 

dd> 
Integrating, and observing that at O, ~7" = °> and (/>=a, we 

obtain 

7© 2= ^/ (cos ^- cosa) - (l) 

The integration of this equation requires the use of elliptic 
integrals, hence only the results will be given here. 



THEORY OF COLUMNS. 



332* 



They are : 
x = — Tr{ — 2aK + 2[E am(2aK) — sin 2 |a: sn{(2a + i)K\sn(2 aK)]} (2) 



y = g{sm ia[-cn{( 2 a + i)K\]\ 

and w^s^J— , 



(3) 



(4) 



where E denotes the elliptic integral of the second kind, and K 
the complete elliptic integral of the first- kind. 

Moreover, for the determination of the load R, we obtain 
from equation (4) 

-iw (5) 



K 



and hence 



*=fV. 



(6) 



From these equations, we can, by using a table of elliptic 
functions, deduce the following results for the coordinates of points 
on the inflectional elastica, for various values of a : 



a 


s 

I 


X 

I 


y 

l 


IO° 


0.00 

0.25 
0.50 


0.0000 
0.2476 
. 4962 


. 0000 
0.0392 
0.0554 


20° 


0.00 

0.25 
0.50 


. 0000 

0.2376 

. 4849 


. 0000 
0.0773 
0. 1079 


30° 


0.00 

0.25 
0.50 


. 0000 
0.2224 
. 4662 


. 0000 

o-H35 
0. 1620 



Moreover, these results agree with those which we obtain by 



H2b APPLIED MECHANICS. 

experiment, and thus we can, by making use of our calculations, 
compute the load required to produce a given elastica, determined 
by the slope at the points of inflection, which, in the case of pin- 
ended columns, are at the ends, and, in the case of columns fixed 
in direction at the ends, are half-way between the middle and the 
ends. 

All this can be done, and can be verified by experiment, 
provided that the load is not so great that any fibre is strained 
beyond the elastic limit of the material, and provided the value of 
l/p is not so small that the curvilinear form is unstable. . 

For smaller values of l/p the only stable form is a straight line, 
and the column does not bend. 

To ascertain the least value of l/p for which a curved form is 
stable, observe that K cannot be less than 71/2, and since this cor- 
responds to one bay, and hence to the case of a pin-ended column, 
we have in that case, by substituting n/2 for K in equation (6), 

7T 2 

R=pE 1 I ) 

and, since I=Ap 2 and 7=^, 

we have for the line of demarcation between the straight and 
curved form in a pin-ended column 



p y a 



(7) 



and for that in the case of a column fixed in direction at the ends 

As an example, if = 10,000 and £1=30,000,000 we should 
find that a pin-ended column would not bend unless l/p were 
greater than 172, and that a column fixed in direction at the ends 



G K 



^' 



UB 

Fig. 239. 



S TR ENG TH OF SHA FTING. 333 

would not bend unless l/p were greater than 344. Columns with 
smaller values of l/p would remain straight when the resultant of 
the load acts along the axis, and no fibre is strained beyond the 
elastic limit. 

§ 212. Strength of Shafting. — The usual criterion for the 
strength of shafting is, that it shall be sufficiently strong to 
resist the twisting to which it is exposed in the transmission of 
power. 

Proceeding in this way, let EF (Fig. 239) be a shaft, AB the 
driving, and CD the following, pulley. 
Then, if two cross-sections be taken 
between these two pulleys, the por- 
tion of the shaft between these two in \? 
cross-sections will, during the trans- 
mission of power, be in a twisted con- 
dition ; and if, when the shaft is at 
rest, a pair of vertical parallel diameters be drawn in these sec- 
tions, they will, after it is set in motion, no longer be parallel, 
but will be inclined to each other at an angle depending upon 
the power applied. Let GH be a section at a distance x from 
O, and let KI be another section at a distance x -j- dx from O. 
Then, if di represent the angle at which the originally parallel 
diameters of these sections diverge from each other, and if r = 
the radius of the shaft, we shall have, for the length of an arc 
passed over by a point on the outside, 

rdi ; 

and for the length of an arc that would be passed over if the 
sections were a unit's distance apart, instead of dx apart, 

rdi _ di 
dx dx 

This is called the strain of the outer fibres of the shaft, as it 
is the distortion per unit of length of the shaft. 



334 APPLIED MECHANICS. 

In all cases where the shaft is homogeneous and symmet- 
rical, if i is the angle of divergence of two originally parallel 
diameters whose distance apart is x> we shall have the strain, 

di i 
v = r — = r-. 

dx x 

This also is the tangent of the angle of the helix. 

A fibre whose distance from the axis of the shaft is unity, 
will have, for its strain, 

di i 

dx x 

A fibre whose distance from the axis of the shaft is p, will have, 
for its strain, 

di i 

Fixing, now, our attention upon one cross-section, GH t we have 
that the strain of a fibre at a distance p from the axis (p varying, 
and being the radius of any point whatever) is 



® 



where - is a constant for all points of this cross-section. 

Hence, assuming Hooke's law, " Ut tensio sic vis" we shall 
have, if C represent the shearing modulus of elasticity, that the 
stress of a fibre whose distance from the axis is p y is 



»-aq-*®> 



which quantity is proportional to p, or varies uniformly from the 
centre of the shaft. 

The intensity at a unit's distance from the axis is 



# 



STRENGTH OF SHAFTING. 335 

and if we represent this by a, we shall have for that at a dis- 
tance p from the axis, 

p = a P . 




Hence we shall have (Fig, 240), that, on a small 
area, 

dA = d P (pd6) = pdpdO, IST^o. 

the stress will be 

pdA = apdA = ap 2 dpdO. 

The moment of this stress about the axis of the shaft is 

ppdA = ap 2 dA = ap^dpdO, 

and the entire moment of the stress at a cross-section is 

afp 2 dA = affp^dpdO = al, 

where / = fp 2 dA is the moment of inertia of the section about 
the axis of the shaft. 

This moment of the stress is evidently caused by, and hence 
must be balanced by, the twisting-moment due to the pull of the 
belt. Hence, if M represent the greatest allowable twisting- 
moment, and a the greatest allowable intensity of the stress at 
a unit's distance from the axis, we shall have 

P 

If f is the safe working shearing-strength of the material 
per square inch, we shall have / as the greatest safe stress per 
square inch at the outside fibre, and hence 

M= -I 
r 

will be the greatest allowable twisting-moment. 



33^ APPLIED MECHANICS. 



For a circle, radius r, 

2 J 2 J l6 

For a hollow circle, outside radius r„ inside radius r„ 

2 ^2^1 ' 

Moreover, if the dimensions of a shaft are given, and the 
actual twisting-moment to which it is subjected, the stress at a 
fibre at a distance p from the axis will be found by means of the 
formula 

The more usual data are the horse-power transmitted and 
the speed, rather than the twisting-moment. 

If we let P = force applied in pounds and R = its leverage 
in inches, as, for instance, when P = difference of tensions of 
belt, and R = radius of pulley, we have 

M=P.R; 

and if HP = number of horses-power transmitted, and 
N = number of turns per minute, then 

HP = P ^ 7CRN ^ 

12 X 33000 ' 

271 N 



EXAMPLE. 

Given working-strength for shearing of wrought-iron as 10000 
lbs. per square inch ; find proper diameter of shaft to transmit 
20-horse power, making 100 turns per minute. 



TRANSVERSE DEFLECTION OF SHAFT. 337 



Angle of Torsion. — From the formula, page 336,/ = -j-. 



combined with 



we have 



/ = ap = Cp-y 
x 



Cp - X ~T 

. Mx 



which gives the circular measure of the angle of divergence of 
two originally parallel diameters whose distance apart is x ; the 
twisting-moment being M, and the modulus of shearing elas- 
ticity of the material, C. 

EXAMPLES. 

i. Find the angle of twist of the shaft given in example 1, § 212, 
when the length is 10 feet, and C = 8500000. 

2. What must be the diameter of a shaft to carry 80 horses-power, 
with a speed of 300 revolutions per minute, and factor of safety 6, break- 
ing shearing-strength of the iron per square inch being 50000 lbs. 

§213. Transverse Deflection of Shafts. — In determining 
the proper diameter of shaft to be used in any given case, we 
ought not merely to consider the resistance to twisting, but 
also the deflection under the transverse load of the belt-pulls, 
weights of pulleys, etc. This deflection should not be allowed 
to exceed y-J-g- of an inch per foot of length. Hence the de- 
flection should be determined in each case. 

The formulae for computing this deflection will not be given 
here, as the methods to be pursued are just the same as in the 
case of a beam, and can be obtained from the discussions on 
that subject. 



33§ APPLIED MECHANICS. 

§ 214. Combined Twisting and Bending. — The most com- 
mon case of a shaft is for it to be subjected to combined twisting 
and bending. The discussion of this case involves the theory 
of elasticity, and will not be treated here ; but the formulae com- 
monly given will be stated, without attempt to prove them until 
a later period. These formulae are as follows : — 
Let M x = greatest bending-moment, 

M 2 = greatest twisting-moment, 

r = external radius of shaft, 

/ = moment of inertia of section about a diameter, 

7r ^ 4 
for a solid shaft / = — , 
4 
f = working-strength of the material = greatest al- 
lowable stress at outside fibre ; 
then 

i°. According to Grashof, 



/= -\W* + fV^ 2 + M* }. (1) 

2°. According to Rankine, 



/ = T j m s + V^ 2 + M 2 2 j . (2) 

§215. Springs. — The object of this discussion is to enable us 
to answer the following three questions : (a) Given a spring, 
to determine the load that it can bear without producing in the 
metal a maximum fibre stress greater than a given amount. 
(b) Given a spring, to determine its displacement (elongation, 
compression, or deflection) under any given load. . (c) Given a 
load P and a displacement d z ; a spring is to be made of a 
given material such that the load P shall produce the displace- 
ment d, , and that the metal shall not, in that case, be subjected 
to more than a given maximum fibre stress. Determine the 
proper dimensions of the spring. 



springs. 339 



There are practically only two cases to be considered as far 
as the manner of resisting the load is concerned. In the first, 
the spring is subjected to transverse stress, and is to be calcu- 
lated by the ordinary rules for beams. In the second, the 
spring is subjected to torsion, and the ordinary rules for re- 
sistance to torsion apply. It is true that in most cases where 
the spring is subjected to torsion there is also a small amount 
of transverse stress in addition to the torsion ; but in a well- 
made spring this transverse stress is of very small amount, and 
we may neglect it without much error. 

We will begin with those cases where the spring is subjected 
to torsion, and for all cases we shall adopt the following nota 
tion : 

P = load on spring producing maximum fibre stress/; 

f = greatest allowable maximum fibre stress for shearing ; 

C = shearing modulus of elasticity ; 

x = length of wire forming the spring ; 
M x = greatest twisting moment under load P; 

L = any load less than the limit of elasticity ; 

M = twisting moment under this load ; 

p = maximum fibre stress under load L ; 

p = distance from axis of wire to most strained fibre ; 

/ = moment of inertia of section about axis of wire ; 

i x = angle of twist of wire under load P\ 
i = angle of twist of wire under load L ; 

V = volume of spring ; 

d 1 = displacement of point where load is applied when load 
isP; 

d = displacement of point where load is applied when load 
is L. 

Then from pages 335 and 337 we obtain the following four 
formulae : 



*=j', to 



/ 


= 


Mx 
CI' 


M x 


= 


£'■ 


h 


= 


M,x 
CI ' 



340 APPLIED MECHANICS. 

(■) 

(3) 

(4) 

These four formulae will enable us to solve all the cases of 
springs subjected to torsion only. Moreover, in the cases 
which we shall discuss under this head, the wire will have either 
a circular or a rectangular section : in the former case we will 
denote its diameter by d, and we shall then have 

r n<? d 

I = and p = — ; 

32 2 

while in the latter case we will denote the two dimensions of 
the rectangle by b and h, respectively, and we shall then have 

We will now proceed to determine the values of P, 6, d x , and 
Fin each of the following four cases, all of which are cases of 
torsion : 

Case i. Simple round torsion wire. — Let AB, the leverage 
of the load about the axis, be R ; then we shall have 

M = IR y M x = PR\ 

and we readily obtain from the formulae (1), (2), (3), and (4) 

M ■ B F=/ ^' (S) 



SPRINGS. 



341 



and from these we readily obtain 






(8) 



CASE 2. Simple rectangular torsion wire. — In this case we 
readily obtain 



j> = i£mVFT#, (9) 

* - *' - Marine' (lo) 

2.&X / , . 



P 






3^ 



V=5jr(PS 1 ). 



(12) 



CASES 3 and 4. Helical springs made of round and of rec- 
tangular wire respectively. — A helical spring may be used either 
in tension or in compression. In either case it is important 
that the ends should be so guided that the pair of equal and 
opposite forces acting at the ends of the spring should act ex- 
actly along the axis of the spring. 

This is of especial importance when the spring is used for 
making accurate measurements of forces, as in the steam-en- 
gine indicator, in spring balances, etc. 

Moreover, it is generally safer, as far as accuracy is con- 
cerned, to use a helical spring in tension rather than in com- 
pression, as it is easier to make sure that the forces act along 



34 2 APPLIED MECHANICS. 

the axis in the case of tension than in the case of compres- 
sion. 

Whichever way the spring is used, however, provided only 
the two opposing forces act along the axis of the spring, the 
resistance to which the spring is subjected is mainly torsion, 
inasmuch as the amount of bending is very slight. 

This bending, however, we will neglect, and will compute 
the spring as a case of pure torsion, the same notation being 
used as before, except that we will now denote by R the radius 





of the spring, and we shall have 

M=ZR, M 1 = RR; 

and now formulae (5), (6), (7), and (8) become applicable to a 
spring made of round wire, and formulae (9) and (10), (11) and 
(12), to one made of rectangular wire. 

We must bear in mind, however, that x denotes the length 
of the wire composing the spring, and not the length of the 
spring. 6 and d 2 now denote the elongations or compressions 
of the spring. 

GENERAL REMARKS. 

By comparing equations (8) and (12), it will be seen 
that if a spring is required for a given service, its volume 
and hence its weight must be 50 per cent greater if made 
of rectangular than if made of round wire. Again, it is 
evident that when the kind of spring required is given f 



springs. 343 



and the values of C and f for the material of which it is tc 
be made are known, the volume and hence the weight of 
the spring depends only on the product PS V and that^as soon 
as P and 6 X are given, the weight of the spring is fixed inde- 
pendently of its special dimensions. If, however, we fix any 
one dimension arbitrarily, the others must be so fixed as to 
satisfy the equations already given. Next, as to the values to 
be used for /"and C, these will depend upon the nature of the 
special material of which the spring is made, and these can 
only be determined by experiment. Confining ourselves now 
to the case of steel springs, it is plain that /and C should be 
values corresponding to tempered steel. 

*As an example, suppose we require the weight of a helical 
spring, which is to bear a safe load of ioooo lbs. with a deflec- 
tion of one inch, assuming C= 12600000 and/= 80000 lbs. 
per sq. in., and as the weight of the steel 0.28 lb. per cubic 
inch. 

From formula (8) we obtain 

2 X 12600000 X 10000 X 1 

v = 80000 x 80000 = 39 ' 4 cu - m - 

Hence the weight of the spring must be (39.4) (0.28) = 1 1 lbs. 
We may use either a single spring weighing 1 1 lbs., or else 
two or more springs either side by side or in a nest, whose com- 
bined weight is n lbs. Of course in the latter case they must 
all deflect the same amount under the portion of the load 
which each one is expected to bear, and this fact must be 
taken into account in proportioning the separate springs that 
compose the nest. 

FLAT SPRINGS. 

Let P, L, V, tf, and d x have the same meanings as before, 

and let 



344 



APPLIED MECHANICS. 



f= greatest allowable fibre stress for tension or compres- 

sion : 
E = modulus of elasticity for tension or compression ; 
/= length of spring; 
M 1 = maximum bending-moment under load P ; 
M = maximum bending-moment under load L. 

Moreover, the sections to be considered are all rectangular, 
and we will let b = breadth and h = depth at the section 
where the greatest bending-moment acts, the depth being 
measured parallel to the load. 

Then if / denote the moment of inertia of the section of 
greatest bending-moment about its neutral axis, we shall have 



/ = 



btf 



12 



We will now consider six cases of flat springs, and will de- 
termine P, d, $ z , and V for each case, and for this purpose we 
only need to apply the ordinary rules for the strength and de- 
flection of beams. 

Case i. Simple rectangular spring, fixed at one end and 
loaded at the other. 



*/ 



bff_ 

I ' 



6 = 4 



M' £' 

III . 
HE' 



(n) 
(24) 
(25) 




•• PS i =\£(bhl)=\£v 



v= 9 fps, 



(26) 



SPRINGS. 



345 



Case 2. Spring of uniform depth and uniform strength, tri- 
wmgular in plan, fixed at one end and loaded at the other. 



° = 1/ 




V = ZyA^)- 



(30) 



Case 3. Spring of uniform breadth and uniform strength, 
parabolic in elevation, fixed at one end and loaded at the 
other. 

f=i/-f, (31) g. 



<y=8 



n 1 






(32) 
(33) 




••P6 t =\ f ~{\bJu)=\ f ~V; 



y=sfA^,). 



(34) 



CASE 4. Compound wagon spring, made up of n simple rec- 
tangular springs laid one above the other, fixed at one end and 
loaded at the other. 



346 



APPLIED MECHANICS. 



Let the breadth be b y and the depth of each separate layer 
be h. Then 



p-i f b JL 



(35) 
(36) 
(37) 




-if 



; J -{nM) = l s f - E V; 



E 



■■ V= 9 y,{P8 J ). 



(38) 



Case 5. Compound spring composed of n triangular springs 
laid one above the other > fixed at one end and loaded at the 
other. 



p= n -fM 




(39) 




6 r l 

nbh* E' 




(40) 




6 >-hE> 




(4t) 




.'. P* x = 


^ E 


Inbhl\ _ 




/. V- 


E 


(^.)- 





(42) 



CASE 6. This case differs from the last in that in order to 
economize material we superpose springs of different lengths, 



SPRINGS. 



347 




and make them of such a shape that by the action of a single 

force at the free end they are bent in arcs of circles of nearly 

or exactly the same radius. 

The force P bends the 

lowest triangular piece AA 

in the arc of a circle. The 

/ 
length of this piece is -. 

In order that the re- 
maining parallelopipedical 
portion may bend into an 
arc of the same circle it is 
necessary that it should have 

acting on it a uniform bending-moment throughout, and this 
is attained if it exerts a pressure at A x upon the succeeding 
spring equal to the force P, and following this out we should 
find that the entire spring would bend in an arc of a circle. 

The values of P, 6, d z , and Fare correctly expressed for 
this case by (39), (40), (41), and (42). 

For any flat springs which are supported at the ends and 
loaded at the middle, or where two springs are fastened to- 
gether, it is easy to compute, by means of the formulae already 
developed, by making the necessary alterations, the quantities 
Ps $ 8 iy and V, and this will be left to the student. 

COILED SPRINGS SUBJECTED TO TRANSVERSE STRESS. 

Three cases of coiled springs will now be given as shown 
in the figures, and the values of P, S, # I} and Fwill be deter- 
mined for each. 

In each of these cases let R be the leverage of the load, 
and let 00 = angle turned through under the load. Then we 
may observe that all the three cases are cases of beams sub- 
jected to a uniform bending-moment throughout their length, 
this bending-moment being LR for load L and PR for load P, 



348 



APPLIED MECHANICS. 



CASES i and 2. Coiled spring, rectangular in section, 
eh' 



P = \f 



l]?L 
~ I2 bh 3 E' 



a - f R1 



(43) 
(44) 
(45) 




/. P5, = \£(bhl) = \£v; 



:. V= i fA^ 1 ). 



(46) 



Case 3. Coiled spring, cir 
tular in section. 

*_(>±IJ?L 
n d> £' 

d £' 




r=4j-A-P6d 



(so) 



TIME OF OSCILLATION OF A SPRING. 

Since in any spring the load producing any displacement 
is proportional to the displacement, it follows that when a 
spring oscillates, its motion is harmonious. 



springs. 349 



Suppose the load on the spring to be P, and hence its nor- 
mal displacement to be <5\. Now let the extreme displacements 
on the two sides of S x be d Q , and the force producing it p, so that 
the actual displacement varies from d z -j- d to d t — d Q , and the 
force acting varies from P -\- p to P — p. 

Now, from the properties of the spring we must have 

Moreover, in the case of harmonic motion the maximum 

WcPr 

value of the force acting is (see p. 104). But the load 

Oscillating is P instead of W, and the extreme displacement is 
d instead of r. 
Hence we have 



;«£<«»*. = *«•*,; (5.) 



g g 



' 6,' 

'•« = /!. (S3) 



Hence the time of a double oscillation 



*=^=**|/A. (S4) 

- a g 



35° APPLIED MECHANICS. 



CHAPTER VII. 

STRENGTH OF MATERIALS AS DETERMINED BY 
EXPERIMENT. 

§ 216. Whatever computations are made to determine the 
form and dimensions of pieces that are to resist stress and 
strain should be based upon experiments made upon the mate- 
rials themselves. 

The most valuable experiments are those made upon pieces 
of the same quality, size, and form as those to which the results 
are to be applied, and under conditions entirely similar to those 
to which the pieces are subjected in actual practice. 

From such experiments the engineer can learn upon what he 
can rely in designing any structure or machine, and this class of 
tests must be the final arbiter in deciding upon the quality of 
material best suited for a given service. An attempt will be made 
in this chapter to give an account of the most important results 
of experiments on the strength of materials, and to explain the 
modes of using the results. 

While the importance of making tests upon full-size pieces, 
and of introducing into the experiments the conditions of 
practice, is pretty generally recognized to-day, nevertheless 
there are some who have not yet learned to recognize the fact 
that attempts to infer the behavior of full-size pieces under 
practical conditions from the results of tests on small models, 
made under conditions which are, as a rule, necessarily, quite 
different from those of practice, are very liable to lead to con- 
clusions that are entirely erroneous. 



GENERAL REMARKS. 35 I 

Such a proceeding is in direct violation of a principle that 
the physicist is careful to observe throughout his work, vis. : 
not to apply the results to cases where the conditions are essentially 
different from those of the experiments. 

When the quality of the material suited for a given 
service is known, tests of the material furnished must be 
made to determine its quality. Such tests, made upon 
small samples, should be of such a kind that there may 
be a clear understanding, as to the quality desired, between 
the maker of the specifications and the producer. Whenever 
possible, standard forms of specimens and standard methods 
of tests should be used. 

The determination of standards is occupying the at- 
tention of the Int. Assoc, for Testing Materials, the British 
Standards Committee, the Am. Soc. for Testing Materials, 
and others. 

To ascertain the quality of the material tensile tests are most 
frequently employed, their objects being to determine the tensile 
strength per square inch, the limit of elasticity, the yield-point, 
the ultimate contraction of area per cent, the ultimate elongation 
per cent in a certain gauged length, and sometimes the modulus 
of elasticity. 

While the standard forms and dimensions will be given later, 
the following general classification of the forms in use will be 
given here, viz. • 

i°. The specimen may be provided with a shoulder at each 
end, having a larger sectional area than the main body of the 
specimen, the section of this being uniform throughout as shown 
in Fig. a, the latter being of so great a length in proportion to the 

diameter that the stretch of i - ,11,, — "" — I 

the specimen is not essentially I << I 

different from what it would FlG# a ' 

be if the section were uniform throughout. The shoulders are, 



35 2 APPLIED MECHANICS. 

of course, the portions of the specimen where the holders (or 

clamps) of the testing-machine are attached. 

2°. In the case of a round specimen of that kind there may 

be a screw-thread on the shoulders as shown in Fig. b. 

In the case of a brittle material, as 

cast-iron or hard steel, it is desirable to 

use a holder with a ball-joint, and to 
Fig. b. ' . - , , , , 

screw the specimen into the holder. 

3°. The specimen may be provided with a shoulder at each 

end, the main body of the specimen being, however, so short 

in proportion to the diameter that the stretch is essentially 

modified. Such a form is shown in Fig. c. 



Fig. c. Ftg. d. 

4°. The specimen may be a grooved specimen as shown in 
Fig. </, where the length of the smallest section is zero. 

5°. The section of the specimen may be uniform through- 
out, the length between the holders being so great in propor- 
tion to the diameter that the stretching of the fibres is not 
interfered with. This form of specimen is shown in Fig. e. 

Assume a specimen of duc- 
l 1 tile material, as mild steel or 

wrought-iron, of the 1st or the 
FlG - *■ 5th shape, subjected to stress 

in the testing-machine, or else by direct weight, and suppose 
that we mark off upon the main body, i.e., the parallel section 
of the specimen, a gauged length of 8 or io inches (preferably 
8 inches), and measure, by means of some form of extensom- 
€ter, the elongations in the gauged length, corresponding to 
the stresses applied ; then plot a stress-strain diagram as shown 
in Fig. /, having stresses per square inch for abscissae, and the 
corresponding strains for ordinates. 



GENERAL REMARKS. 



353 



I 

O .002 













































































'b 






































































































































































20 


00 


60 


00 


10 


X>0 


14000 


18000 


22000 


26000 


30000 


34000 


38000 



LOAD PER SQ. IN. 
Fig. /. 



We shall find that the strains begin by being proportional 
to the stresses, but when a certain stress is reached, called the 
" limit of elasticity " or " elastic limit," shown at A, the strains 
increase more rapidly than the stresses, but the rate of increase 
in the ratio of the strain to the stress is not large until a stress 
is reached called the " yield-point " or " stretch-limit," shown 
at B, which is usually a little larger than the elastic limit ; and 
then the rate of increase of the ratio of strain to stress becomes 
much larger. 

Observe, also, that if a small load be applied to the piece 
under test, and then removed, the deformation or distortion 
caused by the application of the load apparently vanishes, and 
the piece resumes its original form and dimensions on the 
removal of the load ; in other words, no permanent set takes 
place. When the load, however, is increased beyond a certain 
point, the piece under test does not return entirely to its 
original dimensions on the removal of the load, but retains a 
certain permanent set. While permanent set that is easily 
determined begins at or near the elastic limit, and while the 
permanent sets corresponding to stresses greater than the 
elastic limit are much greater than the corresponding recoils, 
and hence form the greater part of the strains corresponding 
to such stresses, nevertheless experiments show that even a 
very small load will often produce a permanent set, and that 
the apparent return of the piece to its original dimensions is, 



354 APPLIED MECHANICS. 

in a number of cases, only due to the want of delicacy in the 
measuring-instruments at our command. 

After the elastic limit and the yield-point have been passed, 
the ratio of the strain to the stress is much greater than before, 
the stretch becomes local, with a local contraction of area, this 
being due to the plasticity of the metal. 

Finally, when the maximum stress is applied, or, in other 
words, the breaking-stress, the behavior is apparently some- 
what different when the piece is subjected to dead weight from 
what it is when in a testing-machine. In the former case, 
when the maximum load is reached, the specimen continues to 
stretch rapidly, without increase in the load, until the specimen 
breaks. 

In the case of the testing-machine, however, the application 
of the maximum load causes, of course, the specimen to 
stretch, but this stretch naturally reduces the load applied, and 
the actual load under which the specimen separates into two 
parts is less, and often very considerably less, than the maxi- 
mum or breaking stress. 

Observe that the terms " breaking-load " and " breaking- 
stress " are always used to mean the " maximum load " and 
"maximum stress " respectively, and are never used to denote 
the load or the stress under which the specimen separates into 
two parts when the latter differs from the former. 

If the stretch of the specimen, as described above, is in any 
way interfered with, the behavior of the specimen will not be 
a proper criterion of the properties of the material ; the per- 
centage contraction of area at fracture will vary with the 
amount of interference with the stretch, and hence with the 
proportions of the specimen ; and the maximum or breaking 
strength will be greater than the real maximum or breaking 
strength per square inch of the material. Hence it follows 
that the 3d and 4th forms of specimen do not indicate cor- 
rectly the quality of the material, furnishing, as they do, 
erroneous values for both breaking-strength and ductility. 



CAST-IRON. 355 



The quantities sought in such tests as those described 
above (with specimens of the 1st, 2d or 5th forms) are, as 
already stated : 

i°. The breaking-strength per square inch of the material; 

2°. The limit of elasticity of the material ; 

3 . The yield-point or stretch-limit of the material; 

4°. The ultimate contraction of area per cent : 

5 . The ultimate elongation per cent in a given gauged 
length ; 

6°. The modulus of elasticity. 

The first gives, of course, the tensile str Lh of the ma- 
terial ; the second and third ought both to be determined, but 
many content themselves with the third alone, since it is much 
easier to obtain. While they are commonly not far apart, it 
is a fact that certain kinds of stress to which the piece may be 
subjected may cause them to become very different from each 
other. The fourth and fifth are the usual ways of measuring 
the ductility of the metal ; and while the fourth is the most 
definite, the fifth is very much employed, and finds favor with 
most iron and steel manufacturers. The sixth is not often 
determined for commercial work, but it is one of the important 
properties of the metal. 

Of these six properties the two most universally insisted 
upon in specifications for material to be used in the construc- 
tion of structures or of machines are ductility, which is 
universally recognized as an all-important matter, and a suit- 
able breaking-strength per square inch, both a lower and an 
upper limit being generally prescribed for this last. 

On the other hand, although cast-iron and hard steel are 
brittle metals when compared with wrought-iron and mild 
steel, nevertheless it is true that the third and fourth forms 
of specimen will show too high results for tensile strength even 
in these materials on account of the interference with the 
stretch of the metal. 



35^ APPLIED MECHANICS. 

§ 217. Cast-iron. — Cast-iron is a combination of iron with 
carbon, the most usual quantity being from 3 to 4 per cent. The 
large amount of carbon which it contains is its distinguishing 
feature, and determines its behavior in most respects. Besides 
carbon, cast-iron contains- such substances as silicon, phosphorus, 
sulphur, manganese, and others. A considerable amount (more 
than 1.37 per cent as stated by Prof. Howe) of silicon forces 
carbon out of combination and into the graphitic form, thus 
lowering the strength. 

Pig-iron is the result of the first smelting, being obtained 
directly from the blast-furnace. The ore and fuel (usually 
coke, though anthracite coal is used to some extent, and some- 
times charcoal) are put into the furnace, together with a flux, 
which is usually limestone, in suitable proportions. The mass 
is brought to a high heat, a strong blast of heated air being intro- 
duced. The mass is thus melted, the fluid metal settling to 
the bottom, while slag, which is the result of the combination 
of the flux with impurities of the ore and fuel, rises to the top. 
The iron is drawn off in the liquid state and run into moulds, 
the result being pig-iron. 

The metal usually contains from 3 to 4 per cent of carbon, 
a part being chemically combined with the iron, and a part in 
the form of graphite. The larger the proportion of combined 
carbon, the whiter the fracture, and the harder and more brittle 
the product, while the larger the proportion of graphite, the darker 
the fracture, and the softer and less brittle the product. That 
which has most of its carbon in combination is called white iron, 
while that which contains a large proportion of graphite is called 
gray cast-iron. 

Pig-iron also contains silicon, sulphur, phosphorus, etc. 
The quantity of the first two can, to a certain extent, be controlled 
in the furnace, but not that of the last, so that if low phosphorus is 
desired, the ore and the fuel used must both be low in phosphorus. 

Gray cast-iron has been, and is sometimes classified in various 



CAST-IRON. 357 



ways, according to the proportions of the combined carbon, and 
of the graphite, but the most modern practice is to sell, buy, and 
specify the iron by means of its chemical composition, and not by 
brands. 

That which contains the largest amount of carbon in mechan- 
ical mixture is, as a rule, soft and fusible, and hence suitable for 
making castings where precision of form is the chief desidera- 
tum, as its fusibility causes it to fill the mould well. For general 
use in construction, where strength and toughness are all-import- 
ant considerations, those grades are required which are neither 
extremely soft nor extremely hard. 

As to the adaptability of cast-iron to construction, it presents 
certain advantages and certain disadvantages. It is the cheapest 
form of iron, It is easy to give it any desired form. It resists 
oxidation better than either wrought-iron or steel. Its com- 
pressive strength is comparatively high when the castings are 
small and perfect. On the other hand, its tensile strength is 
much less than that of wrought-iron, or that of steel, averaging 
in common varieties from 16000 or 17000 to about 26000 
pounds per square inch. It cannot be riveted or welded. It 
is a brittle and not a ductile material, it does not give much 
warning before fracture, and, while the stretch under any 
given load per square inch is decidedly larger than that of 
wrought-iron or steel, its total stretch before fracture is small 
when compared with wrought iron and steel. One of the dif- 
ficulties in the use of cast-iron in construction is its liability to 
initial strains from inequality in cooling. Thus if one part of 
the casting is very thin and another very thick, the thin part 
cools first, and the other parts, in cooling afterwards, cause 
stresses in the thin part. 

The fracture of good cast-iron should be of a bluish-gray 
color and close-grained texture. 

At one time cast-iron was extensively used for all sorts 
of structural work, but it was soon superseded by wrought-iron, 
and later by steel. 

Thus it is no longer used in bridgework, nor for floor- 



358 APPLIED MECHANICS. 

beams of a building, though it is still used to a considerable 
extent for the columns of buildings ; and for this purpose it 
has in its favor the fact that it resists the action of a fire better 
than wrought iron or steel. Thus, in the present day, when 
the steel skeleton construction of buildings is so extensively 
employed, it is very necessary to protect the steel beams and 
columns by covering them with some non-conducting material, 
as, otherwise, they would be liable to collapse in case of fire. 

It is used in cases where the form of the piece is of more 
importance than strength, and also where, on account of its 
form, it would be difficult or expensive to forge ; thus hangers, 
pulleys, gear-wheels, and various other parts of machinery of a 
similar character are usually made of cast-iron, as well as a 
great many other pieces used in construction. It is also used 
where mass and hence weight is an important consideration, 
as in the bed-plates and the frames of machines, etc. 

Malleable Iron. — When a casting, in which toughness is 
required is to be made of a rather intricate form, it is frequently 
the custom to malleableize the cast-iron, i.e., to remove a part 
of its carbon, and the result is — provided the casting is small 
—a product that can be hammered into any desired shape wher* 
c old, but is brittle when hot. 

A list of references to some of the principal experimental 
works on the strength and elasticity of cast iron will be given. 

i°. Eaton Hodgkinson : (a) Report of the Commissioners on the 
Application of Iron to Railway Structures. 

(b) London Philosophical Transactions. 1840. 

(c) Experimental Researches on the Strength and other Prop- 
erties of Cast-Iron. 1846. 

2 . W. H. Barlow : Barlow's Strength of Materials. 

3 . Sir William Fairbairn : On the Application of Cast and Wrought 

Iron to Building Purposes. 
4 . Major Wade (U.S.A.) : Report of the Ordnance Department 

on the Experiments on Metals for Cannon. 1856. 
5°. Capt. T. J. Rodman : Experiments on Metals for Cannon. 
6°. Col. Rosset: Resistenza dei Principali Metalli da Bocchidi Fuoco. 



TENSILE STRENGTH OF CAST-IRON. 359 

7 . Tests of Metals made on the Government Testing Machine at 
Watertown Arsenal, 1887, 1888, 1889, 1890, 1891, 1892, 
1893, l8 94> 1896, 1897, 1898. 

8°. Transactions Am. Soc. Mechl. Engrs. for 1889, p. 187 et seq. 

9 . W. J. Keep: (a) Transverse Strength of Cast-iron. Trans. Am. 
Soc. Mechl. Engrs., 1893. 

(b) Relative Tests of Cast-iron. Trans. Am. Soc. 

Mechl. Engrs., 1895. 

(c) Transverse Strength of Cast-iron. Trans. Am. 

Soc. Mechl. Engrs., 1895. 

(d) Keep's Cooling Curves. Trans. Am. Soc. 

Mechl. Engrs., 1895. 

(e) Strength of Cast-iron. Trans. Am. Soc. 

Mechl. Engrs., 1896. 
io°. Bauschinger: Mittheilungenausdem Mech. Tech. Lab. Miinchen. 

Heft 12, 1885; Heft 15, 1887; Heft 27, 1902; Heft 28. 1902. 
ii°. Tetmajer; Mittheilungen der Materialpnifungsanstalt Zurich. 

Heft 3, 1886; Heft 4, 1890; Hefte 5 and 9, 1896. 
12 . Technology Quarterly. October 1888, page 12 et seq. 
13 . Technology Quarterly. Vol. 7, No. 2; Vol. 10. No. 3. 
14 . Transactions of the American Foundrymen's Association. 
15 . Transactions of the American Society for Testing Materials. 

§ 218. Tensile Strength of Cast-iron. — As the use of 
cast-iron to resist tension has been almost entirely superseded by 
that of wrought-iron and steel, results of tests of full-size pieces 
of cast-iron in tension are not available. Tensile tests, however, 
have been extensively employed to determine the quality ; especially 
so when cast-iron cannon were in use; and tensile tests of cast- 
iron are still made, to a certain extent, for the determination 
of quality. For such tests standard specimens should be used, 
and attempts are being made to reduce their number. 

As the strength that should be attained in such specimens 
will become evident from the Standard Specifications of the 
Am. Soc. for Testing Materials, on page 385 et seq., only a few 
tensile tests will be quoted here, and those, for the purpose of 



360 



APPLIED MECHANICS. 



acquainting the reader with the results of some tensile tests of 
cast-iron. 

About 1840 Eaton Hodgkinson made a few experiments to 
determine the laws of extension of cast-iron, and for this purpose 
used rods 10 feet long and 1 square inch in section. The tables 
of average results are given below. 

These tables show that the ratio of the stress to the strain of 
cast-iron varies with the load, growing gradually smaller as the 
load increases, that with moderate loads the ratio of stress to 



Results of Nine Tensile Tests. Results of Eight Compressive Tests. 



Weights 


Strains in 


Ratio of 


Laid on 


Fractions of 


Stress to 


in Pounds. 


the Length. 


Total Strain. 


IQ53-77 


O. 00c 07 


14050320 


1580.65 


O. OOOI I 


13815720 


2107.54 


O.OOO16 


13597080 


3161.31 


O.OOO24 


132 18000 


4215.08 


0.OOO33 


12936360 


5268.85 


O . 00042 


12645240 


6322 .62 


0.00051 


12377040 


7376.39 


.00061 


12059520 


8430. 16 


0.00072 


1 I 776680 


9483 • 94 


O.OO083 


I 1437920 


io537-7i 


O.OOO95 


11314440 


11591.48 


O.OOI07 


1 084 1 640 


12645.25 


O.OOI2I 


10479480 


13699.83 


O.OOI39 


9855960 


14793.10 


O.OOI55 


9549120 



Weights 


Strains in 


Ratio of 


Laid on 


Fractions of 


Stress to 


in Pounds. 


the Length. 


Total Strain. 


2064.75 


ocoo. 16 


13214400 


4129.49 


0.00032 


12778200 


6194.24 


0.00050 


12434040 


8258.98 


0.00066 


12578760 


10323.73 


0.00083 


12458280 


12388.48 


0.00 100 


12357600 


14453-22 


0.00188 


12245880 


16517.97 


0.00136 


12132240 


18582.71 


0.00154 


12050400 


20647.46 


O.00172 


1 20 1 3680 


24776.95 


0.00208 


11911560 


28906.45 


0.00247 


11679720 


33030 . 80 


0.00295 


11215560 



strain for tension of cast-iron does not differ materially from 
that for compression, and that the difference increases as the 
load becomes greater. The agreement is even closer in the 
case of wrought-iron and steel. 

The gradual decrease of the ratio of stress to strain with the 
increase of load shows that Hooke's law, " Ut tensio sic vis" 
(the stress is proportional to the strain), does not hold true in 



RESULTS OF TESTS. 36 1 

cast-iron. Hence, strictly speaking, cast-iron has no elastic 
limit and no modulus of elasticity, nevertheless we are accustomed 
to call the ratio of the stress to the strain under moderate loads 
the modulus of elasticity of the cast-iron. 

In making specifications intended to secure a good quality 
of cast-iron it is very common to call for a transverse test. 
Indeed the resolutions of the international conferences relative 
to uniform methods of testing recommend, in the case of cast- 
iron: 

(a) Test-pieces to be of the shape of prismatic bars no cm. 
standard length (43") and to have a section of 3 cm. square 
(i".i8), one having an addition on one end, from which cubes 
can be cut for compression tests. 

(b) Three such specimens to be tested for transverse strength. 

(c) The tensile strength to be determined from turned test- 
pieces 20 mm. (o".785) diameter and 200 mm. (7". 8$) long, cut 
from the two ends of the test-pieces broken by flexure. 

(d) The compressive strength to be determined from cubes 
3 cm. (1". 18) on a side cut from the first specimens, pressure 
to be applied in the direction of the axis of the original bar. 

These requirements, while calling for transverse tests, call 
also for tensile and compressive tests. 

T x\ Standard Specifications of the Am. Soc. for Testing 
Maten^ls wil be found on page 385 et seq. 

Inasmuch as the tensile strength has been, and is also made 
the basis of specifications for cast-iron, it is important to con- 
sider what should be attained in this regard. 

For this purpose a few tables of comparatively modern tests 
will be given here, and it will be seen that in the ordinary 
varieties of cast-iron it is easy to secure tensile strengths 
from 16,000 to 25,000 pounds per square inch, and that 
more can be secured by taking proper precautions in the 
manufacture. 

Indeed cast-iron which, when tested in the form of a 
grooved specimen, shows a tensile strength of at least 30,000 






02 



APPLIED MECHANICS. 



pounds per square inch is called gun-iron, this having been a 
requirement of the United States Government, in the days of 
cast-iron cannon, for all cast-iron that was to be used in their 
manufacture. 

The following table is taken from a paper on the Strength 
of Cast-Iron, by Mr. W. J. Keep, published in the Transactions 
of the American Society of Mechanical Engineers for 1896,, 
and it gives the averages of the tensile strengths of the fifteen 
different series of tests recorded in the paper. This table is 
given here merely as an example of the results that can be 
obtained by tension tests upon usual varieties of cast-iron.. 
The table is as follows : 



AVERAGES OF TENSION TESTS OF ROUND BARS. 



No. of Series. 


Area of Section, 
0.375 Sq. In. 


Area of Section 
1. 12 Sq In. 


No. of 
Series. 


Area of Section 
375 Sq. In. 


Area of Section, 
1. 12 Sq. In. 


Breaking Load 
per Sq. Inch. 


Breaking Load 
per Sq. Inch 


Breaking Load 
per Sq Inch. 


Breaking Load 
per Sq. Inch. 


2 
3 

4 

5 
6 

7 
8 


20000 
20580 
25050 
21850 
22425 
25550 
18950 
17700 


15700 
22500 
20450 
19350 
19750 
17200 
17700 
15350 


9 
IO 
II 
12 
13 
14 
15 




14800 






17000 
17500 
213OO 
203OO 
20500 


17700 
I4OOO 
24400 
23525 



The following table of results of tension tests of ordinary 
cast-iron from another source will also be given for the same 
purpose as Mr. Keep's results : 



CAST-IRON. 



3 6 3 



CAST-IRON TENSION. 







^. 








«-v 






c 



•d.S 






a 



•a .5 
















& 









J* 






u 


*iz 




Dimensions. 




6 <«" 


Modulus of 
Elasticity. 


Dimensions. 


■S.S 

c . 




Modulus of 
Elasticity. 




:§># 


■h 






Sfl 


"SS 























s 









s 




T.02 X 1.04 


1.06 


19340 


14857000 


1. 00 X 1. 00 


1. 00 


1 7100 


13333000 


1.03 X 1.02 


1.05 


23910 


15481000 


1. 00 X 1.02 


1.02 


19068 


13680000 


1. 00 X .98 


.98 


21180 


15238000 


1. 00 X 1. 00 


1.00 


18000 


13333000 


1. 00 X .97 


•97 


23227 


15881000 


1. 00 X 1.02 


1.02 


19299 


12057000 


1.0* X 1.06 


1.08 


19830 


14539000 


1.06 X .98 


1.03 


17488 


13249000 


I.M X I.O3 


1.03 


20413 


17632000 


1.00X .98 


•98 


19500 


13250000 


.93 X 1. 00 


•93 


16774 


14337000 


1.02 X 1.02 


1.03 


20747 


14543000 


1. 00 X 1. 00 


1 .00 


18600 


15383000 


T.03 X 1.03 


1.06 


18620 


13434000 


1. 00 X 1. 00 


1. 00 


18000 


16666000 


1. 00 X 1. 00 


1 .00 


18910 


13043000 


1 .00 X 1. 00 


1 .00 


19400 


17911000 










1. 00 X 1. 00 


1. 00 


20950 


15789000 


1. 00 X 1. 00 


1 .00 


19900 


15000000 


1. 00 X i. 00 


1. 00 


22900 


15000000 


1. 00 X 1.02 


1.02 


19594 


13373000 


1 .00 X 1. 00 


1. 00 


22400 


15564000 


1. 01 X I.03 


1.04 


16341 


13 108000 


i. 00 X 1.00 


1 .00 


21300 


15384000 


1. 01 X 1.03 


1.04 


13844 


13640000 


1. 00 X 1 .02 


1.02 


13692 


1 5966000 


1.02 X T.08 


1. 01 


13798 


1 1 840000 


I.O> X I .03 


1.05 


21005 


15075000 


I . OO X I . C2 


1.02 


17647 


12787000 


I 08 * I 23 


i-33 


20600 


1 1 900000 


1.03 X 1 .03 


1.06 


14025 


12568000 


1.0) X 1.03 


1.03 


17067 


12676000 


1 .04 X 1.02 


1.06 


15083 


13466000 


1 oc x 1.03 


1.03 


19900 


12929000 


1.02 X 1.04 


I 06 


16874 


9751900 


.08 X 1.03 


1.02 


16404 


12577000 


1.00 X 1. 00 


1 .00 


20000 1 


13043000 


1. co X 1.02 


1 .02 


16450 


12570000 






' 





Colonel Rosset, of the Arsenal at Turin, made a series of 
experiments upon the influence of the shape of the specimen 
upon the tensile strength. For this purpose he used specimens 
with shoulders ; and, among other tests, he compared the 
strength of the same iron by using specimens the lengths of 
whose smallest parts were respectively 1 metre, 30 millimetres, 
and o millimetres, with the following results : — 



Length of Specimen. 


Tensile Strength, in lbs., per Square Inch. 


1st Cannon. 


2d Cannon. 


3d Cannon. 


1 metre . . 

30 millimetres . 

millimetres . 


31291 
32571 
33993 


25601 

345 62 
36411 


28019 
3001 1 
3001 1 



3^4 APPLIED MECHANICS. 

It will thus be seen that, before we can decide upon the 
quality of cast-iron as affected by the tensile strength, it is 
necessary to know the length of that part of the specimen 
which has the smallest area. Colonel Rosset's tests of cast 
iron were almost entirely confined to high-grade irons, suitable 
to use in cannons. 

He deduced, for mean value of the modulus of elasticity of 
the specimens i metre in length, 20419658 lbs. per square inch : 
this, of course, is a modulus only adapted to these high grades, 
and is not applicable to common cast-iron. 

§ 219. Cast-iron Columns. — In consequence of the high 
compressive strength shown by cast-iron when tested in small 
pieces, and in pieces free from imperfections, it was once 
considered a very suitable material for all kinds of columns. 
Nevertheless, its use for the compression members of bridge 
and roof trusses has been abandoned ; cast-iron having been 
displaced first by wrought-iron and subsequently by steel, 
which is the substance now in use for these purposes. 

The principal reasons for the change are the lack of 
ductility, and the consequent brittleness of cast-iron, that it 
cannot be riveted, and that if it breaks it cannot be easily 
repaired. Cast-iron is, however, used to a very considerable 
extent for the columns of buildings. 

The Gordon, the so-called Euler, and the Hodgkinson 
formulae for the breaking-strength of cast-iron columns, have 
all been given in paragraphs 208, 208a, and 209. They are, 
however, all based upon tests made upon very small columns, 
and do not give results agreeing with the tests of such full-size 
columns as are used in practice. We will next consider, 
therefore, the tests that have been made upon full-size cast-iron 
columns, and the conclusions that are warranted in the light of 
these tests. 

Two sets of tests of cast-iron mill columns have been made 
on the Government testing-machine at Watertown Arsenal; an 
account of these sets of tests is published in their reports of 
1887 and of 1888. 



CAST-IRON COLUMNS. 365 



The first lot consisted of eleven old cast-iron columns, which 
had been removed from the Pacific Mills at Lawrence, Mass., 
during repairs and alterations. 

The second lot consisted of five new cast-iron columns cast 
along with a lot that was to be used in a new mill. 

Of these five, the strength of two was greater than the 
capacity of the testing-machine, hence only three were broken ; 
while in the case of the other two the test was discontinued 
when a load of 800000 lbs. was reached. All the columns con- 
tained a good deal of spongy metal, which of course rendered 
their strength less than it would otherwise have been ; never- 
theless, inasmuch as this is just what is met with in building, 
it is believed that these tests furnish reliable information as to 
what we should expect in practice, and that this information 
is much more reliable than any that can be derived from test- 
ing small columns. 

In all the tests the compressions were measured under a 
large number of loads less than the ultimate strength ; but in- 
asmuch as it is not possible, in the case of cast-iron, to fix any 
limits within which the stress is proportional to the strain, no 
attempt will here be made to compute the modulus of elas- 
ticity. Hence there will be given here a table showing the 
dimensions of the columns tested, their ultimate strengths, 
and, in those cases where they were measured, the horizontal 
and vertical components of their deflections, measured at the 
time when their ultimate strengths were reached, as the Govern- 
ment machine is a horizontal machine. A glance at the table 
will make it evident that we cannot, in the case of such columns, 
rely upon a crushing strength any greater than 25000 or 30000 
lbs. per square inch of area of section. Hence it would seem 
to the writer that, in order to proportion a cast-iron column 
to bear a certain load in a building, we should determine the 
outside diameter in such a way as to avoid an excessive ratio 
of length to diameter ; if this ratio is not much in excess 



3^6 APPLIED MECHANICS. 

of twenty, the extra stress produced by any eccentricity of the 
load due to the deflection of the column will be very slight. 
At the same time see that the thickness of metal is sufficient 
to insure a good sound casting. 

Now, having figured the column in this way, compute the 
outside fibre stress (using the method of § 207) that would 
occur with the loading of the floors assumed to be such as to 
give as great an eccentricity as it is possible to bring upon the 
column. If this distribution of the load is one that is likely 
to occur, then the maximum fibre stress in the column due to 
it ought not to be greatly in excess of 5000 lbs. per square 
inch ; but if it is one which there is scarcely a chance of realiz- 
ing, then the maximum fibre stress under it might be allowed 
to reach 10000 lbs. per square inch. If by adopting the di- 
mensions already chosen these results can be obtained, we may 
adopt them ; but if it is necessary to increase the sectional 
area in order to accomplish them, we should increase it. 

Another matter that should be referred to here is the fact 
that a long cap on a column is more conducive to the produc- 
tion of an eccentric loading than a short one ; hence, that a 
long cap is a source of weakness in a column. 

Other sources of weakness in cast-iron columns are spongy 
places in the casting (which correspond in a certain way with 
knots in wood), and also an inequality in the thickness of the 
two sides of the column, the result of this being the same as 
that of eccentric loading; and it is especially liable to occur in 
consequence of the fact that it is the common practice to cast 
columns on their side, and not on end. The engineer should, 
however, inspect all columns to be used in a building, and reject 
any that have the thickness of the shell differing in different 
parts by more than a very small amount. 

A series of tests of full-size cast-iron columns was made bv 
the Department of Buildings of New York City, under the 
direction of Mr. W. W. Ewing, in December, 1897, upon the:' 

I 



CAST-IRON COLUMNS. 



36; 



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APPLIED MECHANICS. 



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CAST-IRON COLUMNS. 



369 



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370 



APPLIED MECHANICS. 



hydraulic press of the Phoenix Bridge Works. This press 
weighs the load on the specimen plus the friction of the piston, 
the latter being, of course, a variable quantity. Nevertheless 
great pains were taken to determine this friction, and hence 
the results are doubtless substantially correct. 

The results are, it will be seen, similar to those obtained in 
the Watertown tests. The table of results is given below, 
and no farther comments are needed. Subsequently tests 
were made to determine the strength of the brackets. For 
this, however, the reader is referred to the Report itself, or to 
Engineering News of January 20, 1898, and for further details 
of the tests of the columns, to the Report itself, or to Engi- 
neering News of January 13, 1; 



Column 
Number. 


Length, 
Inches. 


Outside 

Diameter, 

Inches. 


Average 
Thick- 
ness, 
Inches. 


Breaking 
Load, 
Lbs. 


Average 
Area Sec- 
tion, 
Sq. In. 


p 

Inches 


I 
? 


Break- 
ing 
Load 
per sq. 
in., lbs. 


I 

II 
B* 
B 4 

5 

6 
XVI 
XVII 

7 
8 


I90.25 
190.25 
190.25 
190.25 
I9O.25 

190.23 
160 
160 
1 20 
I20 


15 
15 
15 
15* 

15 

15 
H to 7 f 

8 

6ft 
6ft 


I 

I* 

II 
Ii 

iA 
1 

*ft 
'ft 

ift 


1356000 

1330000 

I 198000 

1246000 

1632000 

over 

2082000 

651000 

645600 

455200 

474100 


43.98 
49-03 
49-03 
49.48 
50.9I 

51-52 
21.99 
22.87 
17.64 
17-37 


4.96 
4.92 
4.92 
4.98 
4.9I 

4.90 
2.50 

2.48 
I.78 
I.80 


.38.36 
38.67 
38.67 
38.20 
38.75 

38.73 
64.00 
64.52 
67.41 
66.67 


30832 
27126 

24434 
25l8l 

32057 

over 

4041 1 

29604 

28229 

25805 

27236 



60000 



CAST-IRON COLUMNS. 



20000 



■--------------------, -------------- r ----------------- 

.1 Z I__ Z II 

^:E:E:::P:^:Ei::E:=:::E^=E:^:E::::::=:E::g|:i 
:::::::::t::5:::::::;:±-l^:::::::::::::::::::::::? 
ie:::::::::::::?::':::::: :: e: ::::::::::::: :::::::::::: 



Abscissae= length di- 
vided by radius of 
gyration of small- 
est section. 

Ordinates = breaking 
strengths per 
square inch of 
smallest section. 



30 10 



80 90 100 110 120 130 140 160. 



CAST-IRON COLUMNS. 



371 



The cut on page 370 shows a graphical representation of 
the preceding tests of full-size cast-iron columns. 

In Heft VIII (1896) of the Mitt. d. Materialprufungsanstalt 
in Zurich is an account of 296 cast-iron struts tested by Prof. 
Tetmajer; 46 being 3 cm. (1". 18) square will not be men- 
tioned farther. The other 250 were hollow circular, the inside 
diameters being 10 cm. (3". 94), 12 cm. (4". 72), or 15 cm. 
(5".9i); the thicknesses being 1 cm. (o".39) or 0.8 cm. (0^.3 1). 
The lengths varied from 4 m. (13'. 12) to 20 cm. (7". 9). They 
are not the most usual thicknesses of columns for buildings, 
though used to a considerable extent. They might be called 
cast-iron pipe columns. The following table contains all those 
250 cm. (8'.2) long and over, and 1 cm. thick, and one set of 
those 0.8 cm. thick. This will exhibit the character of the results 
for such columns of usual lengths. In computing — the actual 





Thickness 


o". 3 9- 






Thickness 


o". 3 i. 








Outside 




Ultimate 






Outside 




Ultimate 


No. of 


Length, 


Diame- 


I 


Strength, 


No. of 


Length, 


Diame- 


1 


Strength, 
Pounds 


Test. 


Feet. 


ter, 


? 


Pounds 


Test. 


Feet. 


ter, 


p 






Inches. 




per sq. in. 






Inches, 




per sq. in. 


55 


9.84 


4«73 


.77.1 


1 848 1 


207 


13.12 


4.62 


103.9 


11518 


56 


9.84 


4.76 


76.7 


20761 


208 


13.12 


4.61 


103.9 


1 1 660 


57 


8.20 


4.76 


63.4 


28156 


209 


ii. 48 


4.58 


91. 1 


16922 


58 


8.20 


4.78 


63-9 


29862 


210 


11. 48 


4-59 


91.9 


10577 


69 


9.84 


5-63 


64.4 


24174 


211 


9.84 


4.56 


78.9 


194*2 


70 


9.84 


5.61 


64.4 


32564 


212 


9.84 


4.60 


77-7 


19482 


7* 


8.20 


5-65 


53-3 


36546 


213 


8.20 


4-56 


65.4 


31843 


72 


8.20 


5-63 


53-4 


47353 


214 


8.20 


4.61 


64.7 


33 J 33 


86 


9.84 


6.69 


53-2 


32564 


225 


13.12 


5.41 


8 7 .8 


J3 °i 
15216 


87 


9.84 


6.67 


53-3 


34270 


226 


13-12 


5-43 


87.5 


I7623 


88 


8.20 


6.73 


43-9 


44224 


227 


11.48 


5-4 1 


76.7 


22326 


89 


8.20 


6.69 


44.1 


46642 


228 


11.48 


5-39 


77-3 


2I3||B 












229 


9.84 


5.39 


66.4 


23748 












230 


9.84 


5.41 


66.1 


23463 












231 


8.20 


S-4i 


54-8 


381 10 












232 


8.20 


5.41 


54-5 


36688 












243 


13.12 


6.56 


71.8 


22041 












244 


13.12 


6-54 


71.9 


24885 












245 


11.48 


6.56 


62.6 


27729 












246 


11.48 


6.56 


62.5 


28156 










- 


247 


9.84 


6-53 


53-9 


35520 












248 


9.84 


6.54 


53-9 


31853 












249 


8.20 


6.56 


44-8 


41949 












250 


8.20 


6.56 


44.8 


4536; 



372 APPLIED MECHANICS. 

length of the strut has been used, whereas Tetmajer adds to 
this 9 ".84, the thickness of the platforms of the machines, as 
they bore on knife-edges. 

Prof. Bauschinger of Munich made two series of tests of 
full-size cast- and of wrought-iron columns to determine the 
effect of heating them red-hot and sprinkling them with water 
while under load. They were loaded in his testing-machine 
with their estimated safe load as calculated from the formulas. 

For cast-iron, 

19912^ m 

— / 3 ' 
1 -f- 0.0006-5 

For wrought-iron, 

11378,4 

— P f 

I + O.OOOO9 ~s 

where P = safe load (factor of safety five), A = area of section, 
/ = length, p = least radius of gyration, pounds arid inches 
being the units. 

A fire was made in a U-shaped receptacle under the post, 
so arranged that the flames enveloped the post. The tem- 
perature was determined from time to time by means of alloys 
of different melting-points ; and the horizontal and vertical 
components of the deflections were read off on a dial as indi- 
cated by a hand attached to the post by a long wire. The 
post was also examined for cracks or fractures. 

In the 1884 series he tested six cast-iron posts of various 
styles, and three wrought-iron posts, one of them being made 
of channel-irons and plates put together with screw-bolts, one 
of I irons and plates also put together with screw-bolts, and 
one hollow circular. 

The details of the tests will not be given here, but only 
Bauschinger's conclusions. He said: 

That wrought-iron columns, even under the most favorable 



CAST-IROiV COLUMN'S. 373 

adjustment of their ends and of the manner of loading, bend 
so much that they cannot hold their load, sometimes with a 
temperature less than 600 ° Centigrade, and always when they 
are at a red heat ; and this bending is accelerated by sprink- 
ling on the opposite side, even when only the ends of the post 
are sprinkled. 

That under similar circumstances cast-iron posts bend, and 
this bending is increased by sprinkling ; but it does not exceed 
certain limits, even when the post is red for its entire length 
and the stream of water is directed against the middle, and the 
post does not cease to bear its load even when cracks are de- 
veloped by the sprinkling. Only when both ends of a cast-iron 
post are free to change their directions does sprinkling them 
at the middle of the opposite side when they are red make 
them break, but such an unfavorable case of fastening the ends 
hardly ever occurs in practice. 

That the cracks in the columns tested occurred in the 
smooth parts, and not at corners or projections. 

That the result of these tests warns us to be much more 
prudent in regard to the use of wrought-iron in building. If 
posts which are subjected to a longitudinal pressure bend so 
badly when subjected to heat on one side that they lose the 
power of bearing their load, how much more must this be the 
case with wrought-iron beams ; and he urges the importance of 
making more experiments. 

In Heft XV of the Mittheilungen he says that the results 
were criticised in two ways, viz. : Moller claiming that he 
should have used different constants, and Gerber that the 
wrought-iron posts were not properly made. 

Bauschinger therefore concluded to make a new set of tests, 
and for this purpose he had made two cast-iron and five 
wrought-iron columns — the former being carefully cast, but on 
the side, while the wrought-iron ones were made by a bridge 
company of very good reputation, and four of them were 
similar to those made at the time for a new warehouse in 
Hamburg. 



374 APPLIED MECHANICS. 



The tests were made just as before, and the following are 

his conclusions : 

That when wrought-iron posts are as well constructed as 
the two referred to, they resist fire and sprinkling tolerably 
well, though not as well as cast-iron ; but that posts con- 
structed like the other three, even with the fire alone, and 
before the sprinkling begins, get so bent that they can no 
longer hold their load. Good construction requires that the 
rows of rivets shall extend through the entire length of the 
post, and the rivets should be quite near each other ; but the 
tests are not extensive enough to show what are the necessary 
requirements to make wrought-iron posts able to stand fire and 
sprinkling ; in order to know this more experiments are needed. 

In Dingler's Polytechnisches Journal for 1889, page 259 et 
seq., is an article by Professor A. Martens, of Berlin, upcn 
the behavior of cast- and wrought-iron in fires, considering 
especially the burning of a large warehouse in Berlin, and 
advocating the protection of iron-work by covering it with 
cement. He says that there are two series of tests upon 
this subject, one of which is the tests of Bauschinger already 
explained, and the other a set of tests made by Moller and 
Luhmann. 

No detailed account of these tests will be given here, but 
only Moller's conclusions, as stated by Prof. Martens, which 
are as follows: 

i°. With ten cast-iron posts he could not get any cracks 
by sprinkling at a red heat; but it is to be noted that his were 
new posts, while those used in Bauschinger's first series were 
old ones, and that those in Bauschinger's second series, which 
were new and very carefully cast, did not show cracks either. 

2°. He claims that while the cracks would allow the post 
still to bear a centre load, it could not bear an eccentric load 
or a transverse load. 

3 . He claims that the load on a cast-iron post should be 
limited to one which shall not produce sufficient bending to 
bring about a tensile stress anywhere when the post is bent by 
the heat and sprinkling. 



TRANSVERSE STRENGTH OF CAST-IRON 375 

4°. He claims that in either cast- or wrought-iron posts, if 
the ends are not fixed, the ratio of length to diameter should 
not exceed 10, whereas if they are it should not exceed 17; 
also, that there is no such thing as absolute safety from fire 
with iron. 

5°. A covering of cement delays the action of the fire, and 
that therefore such a covering is a protection to the post 
against excessive one-sided heating and cooling. 

6°. Cast-iron is more likely to have at any one section a 
collection of hidden flaws than wrought-iron. 

§ 220. Transverse Strength of Cast-Iron. — At one time 
cast-iron was very largely used for beams and girders in build- 
ings to support a transverse load. Its use for this purpose has 
now been almost entirely abandoned, as it has been superseded 
by wrought-iron and steel. 

A great many experiments have been made on the trans- 
verse strength of cast-iron ; the specimens used in some cases 
being small, and in others large. The records of a great many 
experiments of this kind are to be found in the first four books 
of the list already enumerated in § 217. The details of these 
tests will not be considered here, but an outline will be given 
of some of the main difficulties that arise in applying the results 
and in using the beams. 

Cast-iron is treacherous and liable to hidden flaws ; it is 
brittle. It is also a fact that in casting any piece where the 
thickness varies in different parts, the unequal cooling is liable 
to establish initial strains in the metal, and that therefore 
those parts where such strains have been established have 
their breaking-strength diminished in proportion to the amount 
of these strains. 

In the case of cast-iron also, the ratio of the stress to the 
strain is not constant, even with small loads, and is far from 
constant with larger loads ; also, inasmuch as the compressive 
strength is far greater than the tensile, it follows that, in a 
transversely loaded beam which is symmetrical above and be- 
low the middle, the fibres subjected to tension approach their 



37 6 APPLIED MECHANICS. 

full tensile strength long before those subjected to compression 
are anywhere near their compressive strength. The result of 
all this is, that if a cast-iron beam be broken transversely, and 
the modulus of rupture be computed by using the ordinary 
formula, 

x- My 

we shall find, as a rule, a very considerable disagreement be- 
tween the modulus of rupture so calculated and either the 
tensile or compressive strength of the same iron. Indeed, 
Rankine used to give, as the modulus of rupture for rectangu- 
lar cast-iron beams, 40000 lbs. per square inch, and for open- 
work beams 17000 lbs. per square inch, which latter is about 
the tensile strength of fairly good common cast-iron. 

A great deal has been said and written, and a good many 
experiments have been made, to explain this seeming disagree- 
ment between the modulus of rupture as thus computed, and 
the tensile strength of the iron. Barlow proposed a theory 
based upon the assumption of the existence of certain stresses 
in addition to those taken account of in the ordinary theory of 
beams, but his theory has no evidence in its favor. 

Rankine claimed that the fact that the outer skin is harder 
than the rest of the metal would serve to explain matters, but 
this would not explain the fact that the discrepancy exists in 
the case of planed specimens also. 

Neither Barlow nor Rankine seems to have attempted to 
find the explanation in the fact that the formula 

f= My 

/ / 

assumes the proportionality of the stress to the strain, and 
hence that is less and less applicable the greater the load, and 
hence the nearer the load is to the breaking load. An article 
by Mr. Sondericker in the Technology Quarterly of October, 



TRANSVERSE STRENGTH OF CAST-IRON. 377 

1888, gives an account of some experiments made by him to 
test the theory that " the direct stress, tension, or compression, 
at any point of a given cross-section of a beam, is the same 
function of the accompanying strain, as in the case of the cor- 
responding stress when uniformly distributed," and the results 
bear out the theory very well ; hence it follows that, if we use 
the common theory of beams, determining the stresses as such 
multiples of the strains as they show themselves to be in direct 
tensile and compressive tests, the discrepancies largely vanish, 
and those that are left can probably be accounted for by initial 
stresses due to unequal rate of cooling, and by the skin, or 
by lack of homogeneity. In the same article he quotes the 
results of other tests bearing more or less on the matter, and 
there will be quoted here the table on page 378. 
If, therefore, we wish to make use of the formula 

M=f- 

y 

in calculating the strength of cast-iron beams, we cannot use 
one fixed value of f for all beams made of one given quality 
of cast-iron, but we shall have to use a very varying modulus 
of rupture, varying especially with the form, and also with the 
size of the beam under consideration. Now, in order to do 
this, and obtain reasonably correct results, we need, wherever 
possible, to use values of /that have been deduced from ex- 
periments upon pieces like those which we are to use in prac- 
tice, and under, as nearly as possible, like conditions. 

There are not very many records of such experiments avail- 
able, and, in cases where we cannot obtain them, it will prob- 
ably be best to use a value of / no greatei than the tensile 
strength for complicated forms, and forms ha\ing thin webs. 
For pieces of rectangular or circular section we might probably 
use, for good fair cast-iron, 25000 to 30000 lbs. per square 
inch. 

A few tests of the character referred to have been made in 
the engineering laboratories of the Massachusetts Institute of 



378 



APPLIED MECHANICS. 







Modulus of 






! __5 


Form of 
Beam Sec- 
tion. 


Tensile 

Strength, 

lbs. per Sq. 

In. 


Rupture 

My 

J 1 ' 

lbs. per Sq. 

In. 


Ratio. 


Condition 

of 
Specimen. 


Experimenter. 




I9850 


41320 


2.08 


Turned 


C. Bach.* 


^n^ 


16070 


35500 


2.21 


Turned 


Considere. f 


111111 


34420 


63330 


I.84 


Turned 


Considere. 


\H|r 


24770 


54390 


2.I9 


Turned 


Robinson and Segundo. % 




25040 


46280 


I.85 


Rough 


Robinson and Segundo. 




Mean. 


2.03 












16070 
36270 


29250 
58760 


1.82 
1.62 


Planed 
Planed 


Considere. 




WM 




Considere. 




■ 




19090 


33740 


1-77 
1.74 


Planed 


C. Bach. 








Mean. 






19470 


34000 


1-75 


Planed 


C. Bach. 


m 


31430 


49030 


I.56 


Planed 


Considere. 


HI 


19880 


33860 


I.70 


Planed 


Sondericker. 




24770 


42340 


1. 71 

1.68 
1.68 


Planed 
Rough 


Robinson and Segundo^ 
Robinson and Segundo. 








25040 


42110 




Mean. 






19470 


28150 


1-45 


Planed 


C. Bach. 




16070 


22500 


1.40 


Planed 


Considere. 




31860 


36640 


1. 15 


Planed 


Considere. 








25040 


3I3IO 


1.25 
1. 31 


Rough 


Robinson and Segundo. 






Mean. 






16070 


23780 


1.48 


Planed 


Considere. 


1 


31290 


34730 


1. 11 


Planed 


Considere. 


1 


18050 


24550 


1.36 


Planed 


Sondericker. 


KW5S) 


22470 


26150 


1.16 


Rough 


Burgess and Viel6. § 




Mean. 


1.28 







•See Zeitschrift des Vereines Deutscher Ingenieure, Mar. 3d and xoth, «888. 

+ See Annales des Fonts et Chauss^es, 1885. 

JSee Proceedings Institute of Civil Engineers, Vol 86. 

§ See Proceedings Am. Soc. Mechl. Engrs. 1889, pp. 187 et seq. 



TRANSVERSE STRENGTH OF CAST-IRON. 



379 



Technology, and a brief statement of them will be given here. 
The first that will be referred to here is a series of experiments 
made by two students of the Institute, an account of which is 
given in the Proceedings of the American Society of Mechani- 
cal Engineers for 1889, pp. 187 et seq. 

The object of this investigation was to determine the trans- 
verse strength of cast-iron in the form of window lintels, and 
also the deflections under moderate loads, and from the latter 
to deduce the modulus of elasticity of the cast-iron, and to 
compare it with the modulus of elasticity of the same iron, as 
determined from tensile experiments ; also the tensile strength 
and limit of elasticity of specimens taken from different parts 
of the lintel were determined. 

The iron used was of two qualities, marked P and 5 respec- 
tively. 

The tensile specimens were cast at the same time, and from 
the same run as the lintels. 

Besides this, one of each kind of window lintels was cut up 
into tensile specimens, and the specimens were so marked as to 
show from what part of the lintel they were cut. 

The tables of tests will now be given, and the following ex- 
planation of the symbolism employed. 

P and 5 are used, as already stated, to denote the quality 
of the iron. 

A and B are used to denote, respectively, that the specimen 
was unplaned or planed. 

1, 2, 3, etc., denote the number of the test made on that 
particular kind and condition. 

j_ 

11 



38o 



APPLIED MECHANICS, 



I., II., III., denote that the piece has been taken from a 
lintel, and also from what part, as will easily be seen by the 
sketch on page 379. 

Thus P. B. 3 would signify that the specimen was of quality 
P y had been planed, and was the third test of this class. 

On the other hand, P. B. 3 II., would signify in addition 
that it had been taken from a lintel, and was a piece of one of 
the strips marked II. in the sketch. 

The following is a summary of the breaking-weights per 
square inch of the specimens not cut from the lintels : 



P. A. 1 . 
P. A. 2. 
P. A. 3. 
P. A. 4. 



•23757 
.21423 
.18938 
.21409 



4)855 2 7 
21382 



S. A. 1 
S. A. 2 
S. A. 3 



. .24204 
..25258 
. . 24706 

3)74168 

24723 



P. B. 1 21756 


S. B. 1 29574 


P. B. 3 25207 


S. B. 2 23201 



2)46963 



2)52775 



23482 26388 

The following are the breaking-weights per square inch of 
the specimens cut from the iinteis : — 



P. B. 



5 


I 


19651 




r 6 


I 


29124 


6 


I 


20715 




7 


I 


28372 


9 


I 


21076 




8 


I 


25425 


10 


I 


21483 




3 


II 


24704 


4 


II 


19016 


S. B.j 


4 


II 


29414 


7 


II 


19376 




5 


II 


23610 


11 


II 


22146 




9 


III 


27523 


12 


II 


20552 




10 


III 


18301 


2 


III 


l°594 ( *t£> 




I * 


IV 


19616 


13 


III 


16141 








8 


IV 


I0616 











TRANSVERSE STRENGTH OF CAST-IRON. 



381 



All the window lintels tested were of the form shown in 
the figure, and all were supported at the ends and loaded at 
the middle, the span in every case being 52". From the cut 
it will be seen that the web varied in height, being 4 inches 
high above the flange in the centre, and decreasing to 2.5 inches 
at the ends over the supports. 

The following are the results of the separate tests, where 
tensile modulus of rupture means the outside fibre stress per 
square inch on the tension side, and compressive modulus of 
rupture that on the compression side, both being calculated 
from the actual breaking load by the formula 



/ 



_ My 





Breaking-Load, 
Lbs. 


Tensile Modulus of 


Compressive Modulus 


Mark on Lintel. 


Rupture, 


of Rupture, 




lbs. per Sq. In. 


lbs. per Sq. In. 


P. I 


27220 


26648 


81578 


P. 2 


30520 


29879 


91467 


P. 3 


27200 


26659 


81608 


S. 1 


26750 


26198 


80164 


S. 2 


I9850 


J 9433 


59490 


S. 3 


28670 


28068 


85924 


S. 4 


25120 


24592 


75285 



The second series of experiments was made by two other 
students, and an account of the work is given in the same 
article as the former one. 

The object was to determine the constants suitable to use 
in the formulae for determining the strength of the arms of 
cast-iron pulleys ; and also, incidentally, to determine the hold- 
ing power of keys and set-screws. 

Some old pulleys with curved arms, which had been in use 
at the shops, were employed for these tests. They were all 



382 APPLIED MECHANICS. 

about fifteen inches in diameter, and were bored for a shaft 
I T 3 ¥ inches in diameter. 

Inasmuch as this size of shaft would not bear the strain 
necessary to break the arms, the hubs were bored out to a 
diameter of \\^ inches diameter, and key-seated for a key one- 
half an inch square. 

In order to strengthen the hubs sufficiently, two wrought- 
iron rings were shrunk on them, so as to make it a test of the 
arms and not of the hub. 

The pulley under test is keyed to a shaft which, in its turn, 
is keyed to a pair of castings supported by two wrought-iron I- 
beams, resting upon a pair of jack-screws, by means of which 
the load is applied. A wire rope is wound around the rim of 
the pulley, and leaves it in a tangential direction vertically. 
This rope is connected with the weighing lever of the machine, 
and weighs the load applied. 

In a number of the experiments one arm gave way first, 
and then the unsupported part of the rim broke. 

The breaking-load of the separate pulleys was, of course, 
determined, and then it was sought to compute from this the 
value of / from the formula 

Pxy 



f 



nl 



which is the one most commonly given for the strength of 
pulley arms, and which is based upon several erroneous assump- 
tions, one of which is that the bending-moment is equally 
divided among the several arms. In this formula 

/== moment of inertia of section, 

n = number of arms, 

y = half depth of each arm = distance from neutral axis to 
outside fibre, 

x = length of each arm in a radial direction, 

P =± breaking-load determined by experiment. 



TRANSVERSE STRENGTH OF CAST-IRON. 



383 



The results are given in the following table, the units being 
inches and pounds : 






o o 
£ Sr. 



t»2 
S.I 

H 

3 ** 

co C 

4/ 

o > 



o o 

•a "a 

CO CO 

a c« 



S.S 



II 



Sc 

V 

g* 



•a w 

■i! 

<L> CO 

So 

.O C 
CO O 

.rt co 






! § I 
si S 

•O en 

c 



I 41 u 

•5 ° 
c^o- 

"" 4; co 

*ssa 



8T--1 



bo3 « «U 

S a o-S 

■Mas 

S3 v 3 u 

.3..M.O C 

H O 



. cu co C rt 

o2.2j<t3 
H 



a . 
§ 8 
SI 



•0*0 3 



U u X!- 15 J3-* efZ 

•2-2 Js Js « <= 
XX < < o 



~ J8 

a o 

rt -e' a-S 



a 

13 . 

, Xi 
<U 3 

O 



Sfo 


<N 


<+ 


vH 




3 


O 

CO 


1 


00 


ftj« 


rh 


CO 




«N 





^ 


m 


A 


-4- 


CO 


«N 


00 


CO 




CO 


VO 


«N 


«N 


CO 


CO 


IN 


N 


M 


«N 


•iqSxo^V 3uiJfe3.ig 


888 

vO CO C* 


8 


8 




1 


8 

CO 


O 

8 


8 

CO 






xi^xr^ «N 


«N 


VO 


Tt- 


■«■ 


«N 


rf- 





c = 
.2 « 

en — 

Q I 
«1 



XXX X X 



icf» «ojs» t-b» t-b» 
X X X X 

HS ^ H« $ 



■Hs^lsHS Hs "*" Hs "•* 

XXX X X X X 



X 



X 



X 



sips 
X X 



* Hs -P 



•sauy jo aaqum^j 


xr\ vnvo 


VO 


ir» 


vO 


tn 


ITS 


ir% 


UO 


u-> 


in 




stnjv jo qjSaaT 


>* "■> Th 


"^}" 


H5. 
CO 


u% 


LT» 


% 


-"t 


■* 


ur> 






qr\H jo ssauJjDiqx 




^ 




T-i|ei3 


< 


idfx 


"* 


* 


< 


«H< 




•qn H jo qjp?AV 


Ttrnh CO 

CO 


CO 


CO 


CO 


n- 


CO 




« 


^ 


* 




•aii^i jo sssu^oiqi 


^-C-p 


-B 


H^ 


< 


^ 


-B 


-P 


H5 


-c 


H* 




•aoBji 


rj- co CO 


H 
CO 


CO 


CO 


CO 


CO 


"* 


1? 


CO 


rf 


OJO 


•Xannj jo -uiBia 


Hen 




N 


? 


li^ 




«t 


CO 


ir^ 


« 


" 2 
> 


ns3T 10 jaaxnriNj 


<-i M CO 


-* 


10 


vO 


t^ 


00 


ON 





M 


«N 


<i 



384 



APPLIED MECHANICS. 



In the cases of numbers 5, 7, 8, 9, and 10 some of the arms 
were not broken, the rims were now broken off, and the re- 
maining arms were tested separately, the pull being exerted by 
a yoke hung over the end of the arm, the lower end being at- 
tached to the link of the machine. 

The arms were always placed so that the direction of the 
pull was tangent to the curve of the rim at the end of the arm. 
The actual modulus of rupture was then determined by calcula- 
tion from the experimental results, and is recorded in the 
following table, the units being inches and pounds : — 



Number of 


Dimensions of Sec- 
tion at Fracture : 


Bend of Arm with 


Modulus of 


AverageModulus of 
Rupture for each 


Arms. 


all elliptical. 


or against Load. 


Rupture. 


Pulley. 


5 — 1 


ItV XI 


against 


4539 6 


45396 


7 — 1 


ii x i 


against 


36802 




7 — 2 


i«X| 


against 


39537 




7 — 3 


iHxf 


with 


46407 


40915 


8 — 1 


m x « 


against 


35503 




8 — 2 


i« xff 


against 


36091 




8-3 


iHxtt 


with 


39939 




8-4 


ittxtt 


with 


42469 


38500 


9 — 1 


i**t 


against 


41899 




9 — 2 


iA xfi 


against 


44148 




9 — 3 


i*Xi 


with 


55442 


47163 


10 — 1 


if. x« 


against 


54743 




10 — 2 


*«x« 


against 


50943 




10 — 3 


i«x« 


against 


38605 




10 — 4 


ifXtf 


with 


55229 


49880 



Total... 
Average 



663153 
44210 



STANDARD SPECIFICATIONS FOR CAST-IRON 385 



STANDARD SPECIFICATIONS FOR CAST-IRON, OF THE AMERICAN 
SOCIETY FOR TESTING MATERIALS. 

The standard specifications for cast-iron, of the American 
Society for Testing Materials, contain specifications for i° 
Foundry Pig-iron, 2 Gray Iron Castings, 3 Malleable Iron 
Castings, 4 Locomotive Cylinders, 5 Cast-iron Pipe and Special 
Castings, 6° Cast-iron Car-wheels. Of these, i°, 2 , and 4 will 
be quoted in full, and extracts will be given from 5 . For the 
remainder see the proceedings of the Society. 

AMERICAN SOCIETY FOR TESTING MATERIALS. 
SPECIFICATIONS FOR FOUNDRY PIG-IRON. 

Analysis. 
It is recommended that all purchases be made by analysis. 

Sampling. 

In all contracts where pig-iron is sold by chemical analysis, each 
car load, or its equivalent, shall be considered as a unit. At least one 
pig shall be selected at random from each four tons of every car load, 
and so as to fairly represent it. 

Drillings shall be taken so as to fairly represent the fracture-surface 
of each pig, and the sample analysed shall consist of an equal quantity 
of drillings from each pig, well mixed and ground before analysis. 

In case of disagreement between buyer and seller, an independent 
analyst, to be mutually agreed upon, shall be engaged to sample and 
analyze the iron. In this event one pig shall be taken to represent 
every two tons. 

The cost of this sampling and analysis shall be borne by the buyer 
if the shipment is proved up to specifications, and by the seller if other- 
wise. 

Allowances and Penalties. 

In all contracts, in the absence of a definite understanding to the 
contrary, a variation of 10 per cent in silicon, either way, and of 0.02 
sulphur, above the standard, is allowed. 

A deficiency of over 10 per cent and up to 20 per cent, in the silicon, 
subjects the shipment to a penalty of 4 per cent of the contract price. 



386 



APPLIED MECHANICS. 



Base Analysis of Grades. 

In the absence of specifications, the following numbers, known to 
the trade, shall represent the appended analyses for standard grades 
of foundry pig-irons, irrespective of fracture, and subject to allowances 
and penalty as above: 



Grade. 


PerCent 
Silicon. 


Per Cent 

Sulphur 

(Volumetric). 


Per Cent 

Sulphur 

(Gravimetric). 


No i . . . 
N \ 2' . . . 
No 3 . 
No 4 . . . 


2-75 
2.25 

i-75 
1.25 


0-035 
O.045 
0-055 
0.065 


O.045 
0-055 
0.065 
0.075 



PROPOSED SPECIFICATIONS FOR GRAY IRON CASTINGS. 

Process of Manufacture. 

Unless furnace iron is specified, all gray castings are understood to 
be made by the cupola process. 

Chemical Properties. 

The sulphur contents to be as follows: 

Light castings not over o . 08 per cent. 

Medium castings . . . . " " 0.10 " " 
Heavy castings Ci " 0.12 " " 

Definition. 

In dividing castings into light, medium, and heavy classes, the 
following standards have been adopted : 

Castings having any section less than J of an inch thick shall be 
known as light castings. 

Castings in which no section is less than 2 ins. thick shall be known 
as heavy castings. 

Medium castings are those not included in the above definitions. 
Physical Properties. 

Transverse Test. The minimum breaking-strength of the "Arbi- 
tration Bar " under transverse load shall not be under: 

Light castings 2500 lbs. 

Medium castings 2900 " 

Heavy castings 3300 " 



STANDARD SPECIFICATIONS FOR CAST-IRON. 



387 



In no case shall the deflection be under .10 of an inch. 

Tensile Test. Where specified, this shall not run less than: 

Light castings 18000 lbs. per square inch. 

Medium castings .... 21000 " " " " 
Heavy castings 24000 " ' ■ " ' ' 

The " Arbitration Bar" and Methods of Testing. 

The quality of the iron going into castings under specification 
shall be determined by means of the "Arbitration Bar." This is 
a bar i\ ins. in diameter and 15 ins. long. It shall be prepared as. 
stated further on and tested transversely. The tensile test is not 
recommended, but in case it is called for, the bar as shown in Fig. i, 
and turned up from any of the broken pieces of the transverse test,, 
shall be used. The expense of the tensile test shall fall on the purchaser. 

Two sets of two bars shall be cast from each heat, one set from the 
first and the other set from the last iron going into the castings. Where 



Pattern 





<-l)£-D.-> 



f^^~^| 






the heat exceeds twenty tons, an additional set of two bars shall be 
cast for each twenty tons or fraction thereof above this amount. In 
case of a change of mixture during the heat, one set of two bars shall 
also be cast for every mixture other than the regular one. Each set 
of two bars is to go into a single mold. The bars shall not be rumbled 
or otherwise treated, being simply brushed off before testing. 



388 APPLIED MECHANICS. 

The transverse test shall be made on all the bars cast, with supports 
12 ins. apart, load applied at the middle, and the deflection at rupture 
noted. One bar of every two of each set made must fulfill the re- 
quirements to permit acceptance of the castings represented. 

The mold for the bars is shown in Fig. 2 (not shown here). The 
bottom of the bar is T V of an inch smaller in diameter than the top, 
to allow for draft and for the strain of pouring. The pattern shall not 
be rapped before withdrawing. The flask is to be rammed up with 
green molding-sand, a little damper than usual, well mixed and put 
through a No. 8 sieve, with a mixture of one to twelve bituminous 
facing. The mold shall be rammed evenly and fairly hard, thoroughly 
dried and not cast until it is cold. The test-bar shall not be removed 
from the mold until cold enough to be handled. 

Speed of Testing. 

The rate of application of the load shall be thirty seconds for a 
deflection of .10 of an inch. 

Samples for Chemical Analysis. 

Borings from the broken pieces of the " Arbitration Bar" shall 
be used for the sulphur determinations. One determination for each 
mold made shall be required. In case of dispute, the standards of 
the American Foundrymen's Association shall be used for comparison. 

4 Finish. 

Castings shall be true to pattern, free from cracks, flaws, and ex- 
cessive shrinkage. In other respects they shall conform to whatever 
points may be specially agreed upon. 

Inspection. 

The inspector shall have reasonable facilities afforded him by the 
manufacturer to satisfy him that the finished material is furnished in 
accordance with these specifications. All tests and inspections shall, 
as far as possible, be made at the place of manufacture prior to ship- 
ment. 



STANDARD SPECIFICATIONS FOR CAST-IRON. 389 



SPECIFICATIONS FOR LOCOMOTIVE CYLINDERS. 

Process of Manufacture. 
Locomotive cylinders shall be made from good quality of close- 
grained gray iron cast in a dry sand mold. 

Chemical Properties. 
Drillings taken from test-pieces cast as hereafter mentioned shall 
conform to the following limits in chemical composition: 

Silicon from 1.25 to 1 . 75 per cent 

Phosphorus not over .9 " ' ' 

Sulphur " " .10 " " 

Physical Properties. 

The minimum physical qualities for cylinder iron shall be as 
follows : 

The " Arbitration Test-Bar," ij ins. in diameter, with supports 
12 ins. apart shall have a transverse strength not less than 3000 lbs., 
centrally applied, and a deflection not less than 0.10 of an inch. 

Test-Pieces and Method of Testing. 

The standard test shall be ij ins. in diameter, about 14 ins. long, 
cast on end in dry sand. The drillings for analysis shall be taken 
from this test-piece, but in case of rejection of the manufacturer shall 
have option of analyzing drillings from the bore of the cylinder, upon 
which analysis the acceptance or rejection of the cylinder shall be 
based. 

One test-piece for each cylinder shall be required. 

Character of Castings. 
Castings shall be smooth, well cleaned, free from blow-holes, shrink- 
age cracks, or other defects, and must finish to blue-print size. 

Each cylinder shall have cast on each side of saddle manufacturer's 
mark, serial number, date made, and mark showing order number. 

• 
Inspector. 

The inspector representing the purchaser shall have all reasonable 
facilities afforded to him by the manufacturer to satisfy himself that the 
finished material is furnished in accordance with these specifications. 
All tests and inspections shall be made at the place of the manufacturer. 



39° APPLIED MECHANICS, 

CAST-IRON PIPE AND SPECIAL CASTINGS. 

This specification is divided into the following sections, viz.: i° 
Description of Pipes, 2 Allowable Variation in Diameter of Pipes and 
Sockets, 3 Allowable Variation in Thickness, 4 Defective Spigots may 
be Cut, 5 Special Castings, 6° Marking, 7 Allowable Percentage of 
Variation in Weight, 8° Quality of Iron, 9 Tests of Material, io° Cast- 
ing of Pipes, n° Quality of Castings, 12-° Cleaning and Inspection, 13° 
Coating, 14 Hydrostatic Test, 15 Weighing, 16 Contractor to Furnish 
Men and Materials, 17 Power of Engineer to Inspect, i8 p Inspector 
to Report, 19 Castings to be Delivered Sound and Perfect, 20 Defi- 
nition of the Word Engineer. 

Of these, only sections 8° and 9 will be quoted here, as follows: 

Quality of Iron. , 

Section 8. All pipes and special castings shall be made of cast- 
iron of good quality, and of such character as shall make the metal 
of the castings strong, tough, and of even grain, and soft enough to 
satisfactorily admit of drilling and cutting. The metal shall be made 
without any admixture of cinder-iron or other inferior metal, and shall 
be remelted in a cupola or air furnace. 

1 

Tests of Material. 

Section 9. Specimen bars of the metal used, each being 26 inches 
long by 2 inches wide and 1 inch thick, shall be made without charge 
as often as the engineer may direct, and, in default of definite instruc- 
tions, the contractor shall make and test at least one bar from each heat 
or run of metal. The bars, when placed flatwise upon supports 24 
inches apart and loaded in the centre, shall for pipes 12 inches or less 
in diameter support a load of 1900 pounds and show a deflection of 
not less than .30 of an inch before breaking, and for pipes of sizes larger 
than 1 2 inches shall support a load of 2000 pounds and show a deflection 
of not less than .32 of an inch." The contractor shall have the right to 
make and break three bars from each heat or run of metal, and the test 
shall be based upon the average results of the three bars. Should 
the dimensions of the bars differ from those above given, a proper 
allowance therefor shall be made in the results of the tests. 



WR O UGHT-IR ON. 3 9 1 



§ 221. Wrought-iron. — Wrought-iron is obtained by melt- 
ing pig-iron in contact with iron ore, oxidizing, and burning out, 
as far as may be, the carbon, the phosphorus, and the silicon. 
In many cases, however, the charge consists largely of wrought- 
iron or steel scrap, and cast-iron borings. 

The process is commonly carried on in a puddling furnace, 
where an oxidizing flame is passed over the melted pig-iron. 

As the heat is not sufficiently intense to melt the wrought- 
iron produced, the metal is left in a plastic condition, full of 
bubbles and holes, which contain considerable slag. It is then 
squeezed, and rolled or hammered, to eliminate, as far as possible, 
the slag, and to weld the iron into a solid mass. 

The result of this first rolling is known as muck-bar, and must 
be "piled," heated, and rolled or hammered at least once more 
before it is suitable for use in construction. 

In making the piles, while muck-bar is sometimes used 
exclusively, a considerable part, and often the greater part, is 
made of scrap. 

Wrought-iron is thus, throughout its manufacture, a series 
of welds. Moreover, wherever slag is present, these welds cannot 
be perfect. It is also subject to the impurities of the cast-iron 
from which it is made. Thus, the presence of sulphur makes 
it red-short, or brittle when hot; and the presence of phosphorus 
makes it cold-short, or brittle when cold. 

It cannot, like cast-iron, be melted and run into moulds; 
but it can be easily welded by the ordinary methods 

Wrought-iron is much more capable of bearing a tensile or 
transverse stress than cast-iron: it is tougher, it stretches more, 
and gives more warning before fracture. At one time cast-iron 
was the principal structural material, but it was soon displaced 
by wrought-iron, which became the principal metal used in 
construction, but now, since the modern methods of steel-making 
supply a more homogeneous product at a cheaper price, wrought- 
iron has been superseded by mild steel in most pieces used in 
construction. 



392 APPLIED MECHANICS. 

Wrought-iron is also expected to withstand a great many 
trials that would seriously injure cast-iron: thus, two pieces 
of wrought-iron are generally united together by riveting; the 
holes for the rivets have to be punched or drilled, and then the 
rivets have to be hammered; the entire process tending to injure 
the iron. Wrought-iron has to withstand flanging, and is liable 
to severe shocks when in use; as, for instance, those that occur 
from the changes of temperature in the different parts of a steam- 
boiler. 

The following references to a large number of tests of wrought- 
iron will be given : 

i°. Eaton Hodgkinson: (a) Report of Commissioners on the Applica- 
tion of Iron to Railway Structures. 
(b) London Philosophical Transactions. 1840. 

2 . William H. Barlow : Barlow's Strength of Materials. 

3 . Sir William Fairbairn : On the Application of Cast and Wrought 
Iron to Building Purposes. 

4 . Franklin Institute Committee: Report of the Committee of the 
Franklin Institute. In the Franklin Institute Journal of 

1837. 

5 . L. A. Beardslee, Commander U.S.N. : Experiments on the Strength 
of Wrought-iron and of Chain Cables. Revised and enlarged 
by William Kent, M.E., or Executive Document 98, 45th 
Congress, as stated below. 

6°. David Kirkaldy: Experiments on Wrought-iron and Steel. 

7 . G. Bouscaren: Report on the Progress of Work on the Cincinnati 
Southern Railway, by Thomas D. Lovett. Nov. 1, 1875. 

8°. Tests of Metals made at Watertown Arsenal. Of these the first 
two volumes were published before 1881, and since that 
time one volume has been published every year. Nearly all 
of them contain tests of wrought-iron and a great many of 
them contain tests of full-size pieces of wrought-iron. 

9 . A, Wohler: (a) Die Festigkeits versuche mit Eisen und Stahl. 

(b) Strength and Determination of the Dimensions of Structures 



TENSILE STRENGTH OF WROUGHT-IRON. 393 

of Iron and Steel, by Dr. Phil. Jacob J. Weyrauch. Translated 

by Professor Dubois. 
io°. Technology Quarterly, Vol. VII. No. 2, Vol. VIII. No. 3, Vol. 

IX. Nos. 2 and 3, and Vol. X. No. 4. 
ii°. Mitt, der Materialprlifungsaustalt in Zurich. 
12 . Mitt, aus dem Mech. Tech. Lab. in Berlin. 
13° Mitt, aus dem Mech. Tech. Lab. in Mtinchen. 

§ 222. Tensile Strength of Wrought-iron.— About the 
year 1840 was published the report of the Commission appointed 
by the British Government to investigate the application of iron 
to railway structures. While a number of tests of iron had been 
previously made, this work may properly be regarded as having 
been the first investigation of the kind that was at all thorough. 
At that time cast-iron was the metal most used in construction, 
and hence the greater part of the work of the Commission was 
devoted to a study of that metal. They made, however, a number 
of tests of wrought-iron, which, though they were of the greatest 
value at the time, and still have some value, will not be quoted 
here. 

At about that time the use of wrought-iron began to increase 
at a rapid rate, the necessary appliances were introduced to roll 
it into I beams, channel-irons, angle-irons, and other shapes, 
and it began to displace cast-iron for one after another purpose 
until it came to be the metal most extensively used in construction, 
both in the case of structures and machines. 

At first the chief desideratum was assumed to be that it 
should have a high tensile strength, and scarcely any attention 
was paid to its ductility. 

About 1865, however, engineers began to realize that duc- 
tility is an all-important property of a metal to be used in 
construction, and that this is not necessarily and not generally 
obtainable with a very high tensile strength. The most 



394 APPLIED MECHANICS. 

prominent advocate, at that time, of the importance of duc- 
tility was David Kirkaldy, who published a book, entitled 
" Experiments on Wrought Iron and Steel," containing the 
results of his tests down to 1866. 

In the early part of his book will be found a summary of 
what had been done by earlier experimenters in this line. 

Kirkaldy tested a large number of English irons, determin- 
ing both their breaking-strengths and their ductility. 

In the light of the results obtained by him, he proceeded 
to draw up his famous sixty-six conclusions. 

These sixty-six conclusions will not be quoted here, but 
the following statement will be made regarding the main 
results of his work : 

1°. He proved that the results obtained by testing grooved 
specimens (or specimens of such form as to interfere with the 
flow of the metal while under test) did not indicate correctly 
the quality of the metal, but that such specimens should be 
used as did not interfere with the flow of the metal. 

2°. He advocated, with all the earnestness of which he was 
capable, the conclusion that it was of the greatest importance 
that all wrought-iron used in construction should have a good 
ductility, and, in his tests, he adopted five different methods 
of measuring ductility. 

These methods are : i°. Contraction of area at fracture per 
cent ; 2°. Ultimate elongation per cent ; 3 . Breaking-strength 
per square inch of fractured area ; 4 . Contraction of stretched 
area per cent, i.e., the contraction of area attained when the 
maximum load is first reached; 5 . Breaking-weight per square 
inch of stretched area. Of these only two are. used at the present 
time, the first and second, and they serve as measures of 
ductility. These two are the principal conclusions from Kir- 
kaldy's tests, though he cites a great many more, one of the 
principal of them being his conclusion regarding so-called cold 
crystallization, which will be mentioned later. 



SPECIFICATIONS FOR WROUGHT-IRON. 



395 



Tests of the tensile strength of wrought-iron may be divided 
into two classes : i° those made mainly for the purpose of deter- 
mining the quality of the material, and 2° those made upon such 
full-size pieces as are used in practice to resist tension. 

The tests of the first class are made upon small specimens, 
and, in order that the results may be comparable, the use of 
standard forms and dimensions is, generally, a desideratum. 
The specifications for wrought-iron of the American Society for 
Testing Materials will be given first, as they refer to the kind 
of wrought-iron that is in most common use, and then some 
other tensile tests of various kinds of wrought-iron in small pieces 
will be given. Subsequently tests of wrought-iron eye-bars will 
be quoted. 

AMERICAN SOCIETY FOR TESTING MATERIALS. 
SPECIFICATIONS FOR WROUGHT-IRON. 



Process of Manufacture. 

i. Wrought-iron shall be made by the puddling process or rolled 
from fagots or piles made from wrought-iron scrap, alone or with 
muck-bar added. 

Physical Properties. 

2. The minimum physical qualities required in the four classes of 
wrought-iron shall be as follows : 





Stay-bolt 
Iron. 


Merchant 

Iron. 

Grade "A." 


Merchant 

Iron. 

Grade "B." 


Merchant 

Iron, 

Grade "CV 


Tensile strength, pounds 

per square inch . 
Yield -point, pounds per 

square inch 
Elongation, per cent in 8 

inches 


46000 

25000 

28 


50000 

25000 

25 


48000 

25000 

20 


48000 

25000 

20 



3. In sections weighing less than 0.654 pound per lineal foot, the 
percentage of elongation required in the four classes specified in para- 



39^ APPLIED MECHANICS, 

graph No. 2 shall be 12 per cent., 15 per cent., 18 per cent., and 
21 per cent., respectively. 

4. The four classes of iron when nicked and tested as described in 
paragraph No. 9 shall show the following fracture: 

(a) Stay-bolt iron, a long, clean, silky fibre, free from slag or dirt 
and wholly fibrous, being practically free from crystalline spots. 

(b) Merchant iron, Grade "A," a long, clean, silky fibre, free from 
slag or dirt or any course crystalline spots. A few fine crystalline 
spots may be tolerated, provided they do not in the aggregate exceed 
10 per cent of the sectional area of the bar. 

(c) Merchant iron, Grade "B," a generally fibrous fracture, free 
from coarse crystalline spots. Not over 10 per cent of the fractured 
surface shall be granular. 

(d) Merchant iron, Grade "C," a generally fibrous fracture, free 
from coarse crystalline spots. Not over 15 per cent of the fractured 
surface shall be granular. 

5. The four classes of iron, when tested as described in paragraph 
No. 10, shall conform to the following bending tests: 

(e) Stay-bolt iron, a piece of stay-bolt iron about 24 inches long, 
shall bend in the middle through 180 flat on itself, and then bend in 
the middle through 180 flat on itself in a plane at a right angle to 
the former direction without a fracture on outside of the bent 
portions. Another specimen with a thread cut over the entire length 
shall stand this double bending without showing deep cracks in the 
threads. 

(/) Merchant iron, Grade "A," shall bend cold 180 flat on itself, 
without fracture on outside of the bent portion. 

(g) Merchant iron, Grade "B," shall bend cold 180 around a 
diameter equal to the thickness of the tested specimen, without fracture 
on outside of bent portion. 

(h) Merchant iron, Grade "C," shall bend .cold 180 around a 
diameter equal to twice the thickness of the specimen tested, without 
fracture on outside of the bent portion. 

6. The four classes of iron when tested as described in paragraph 
No. 11, shall conform to the following hot bending tests: 

(i) Stay-olt iron, shall bend through 180 flat on itself, without 



SPECIFICATIONS FOR WROUGHT-IRON. 39? 

showing cracks or flaws. A similar specimen heated to a yellow heat 
and suddenly quenched in water between 8o° and 90 F. shall bend, 
without hammering on the bend, 180 flat on itself, without showing 
cracks or flaws. 

(j) Merchant iron, Grade "A," shall bend through 186 flat on 
itself, without showing cracks or flaws. A similar specimen heated 
to a yellow heat and suddenly quenched in water between 8o° and 
oo° F. shall bend, without hammering on the bend, 180 flat on itself, 
without showing cracks or flaws. A similar specimen heated to a bright- 
red heat shall be split at the end and each part bent back through an 
angle of 180 . It will also be punched and expanded by drifts until 
a round hole is formed whose diameter is not less than nine-tenths of 
the diameter of the rod or width of the bar. Any extension of the 
original split or indications of fracture, cracks, or flaws developed by 
the above tests will be sufficient cause for the rejection of the lot rep- 
resented by that rod or bar. 

(k) Merchant iron, Grade "B," shall bend through 180 flat on 
itself, without showing cracks or flaws. 

(/) Merchant iron, Grade "C," shall bend sharply to a right angle, 
without showing cracks or flaws. 

7. Stay-bolt iron shall permit of the cutting of a clean sharp thread 
and be rolled true to gauges desired, so as not to jam in the threading 
dies. 

Test Pieces and Methods of Testing. 

8. Whenever possible, iron shall be tested in full size as rolled, to 
determine the physical qualities specified in paragraphs Nos. 2 and 3, 
the elongation being measured on an eight inch (8") gauged length. 
In flats and shapes too large to test as rolled, the standard test specimen 
shall be one and one-half inches (ij") wide and eight inches (8") 
gauged length. 

In large rounds, the standard test specimen of two inches (2") 
gauged length shall be used; the center of this specimen shall behalf- 
way between the center and outside of the round. Sketches of these 
two standard test specimens are as follows: 



39* 



APPLIED MECHANICS. 




rtf**- 



! ! 5 I 



1 



Parallel Section_ 
not less than 9 " 






3 



|e 18-about — *| 



Piece to be of Same Thickness as the Plate. 



9. Nicking tests shall be made on specimens cut from the iron as 
rolled. The specimen shall be slightly and evenly nicked on one side 
and bent back at this point through an angle of 180 by a succession of 
light blows. 

10. Cold bending tests shall be made on specimens cut from the 
bar as rolled. The specimen shall be bent through an angle of 180 
by pressure or by a succession of light blows. 

11. Hot bending tests shall be made on specimens cut from the 
bar as rolled. The specimens, heated to a bright red heat, shall be 
bent through an angle of 180 by pressure or by a succession of light 
blows and without hammering directly on the bend. 

If desired, a similar bar of any of the four classes of iron shall be 
worked and welded in the ordinary manner without showing signs of 
red shortness. 

12. The yield-point specified in paragraph No. 2 shall be deter- 
mined by the careful observation of the drop of the beam or halt in 
the gauge of the testing-machine. 



tests of commander eeardslee. 399 

Finish. 

13. All wrought-iron must be practically straight, smooth, free 
from cinder spots or injurious flaws, buckles, blisters or cracks. 

In round iron, sizes must conform to the Standard Limit gauge 
as adopted by the Master Car Builders' Association in November, 
1883. 

Inspection. 

14. Inspectors representing the purchasers shall have all reason- 
able facilities afforded them by the manufacturer to satisfy them that 
the finished material is furnished in accordance with these specifications. 
All tests and inspections shall be made at the place of manufacture 
prior to shipment. 



TESTS OF COMMANDER BEARDSLEE. 

One of the most valuable sets of tests of wrought-iron is that 
obtained by committees D, H, and M of the Board appointed 
by the United States Government to test iron and steel; the 
special duties of these committees being to test such iron as would 
be used in chain-cable, and the chain-cable itself. The chairman 
of these three committees, which were consolidated into one, was 
Commander L .A. Beardslee of the United States Navy. The 
full account of the tests is to be found in Executive Document 
98, 45th Congress, second session; and an abridged account of 
them was published by William Kent, as has been already 
mentioned. 

The samples of bar-iron tested were round, and varied from 
one inch to four inches in diameter. 



J no APPLIED MECHANICS. 

Certain conclusions which they reached refer to all kinds 
of wrought-iron, and will be given here before giving a table of 
the results of the tests. 

i°. Kirkaldy considers the breaking-strength per square 
inch of fractured area as the main criterion by which to deter- 
mine the merits of a piece of iron or steel. Commander 
Beardslee, on the other hand, thinks that a better criterion is 
what he calls the "tensile limit;" i.e., the maximum load the 
piece sustains divided by the area of the smallest section when 
that load is on, i.e., just before the load ceases to increase in 
the testing-machine. 

2°. Kirkaldy had already called attention to the fact that 
the tensile strength of a specimen is very much affected by its 
shape, and that, in a specimen where the shape is such that 
the length of that part which has the smallest cross-section is 
practically zero (as is the case when a groove is cut around 
the specimen), the breaking-strength is greater than it is when 
this portion is long ; the excess being in some cases as much 
as 33 per cent. 

Commander Beardslee undertook, by actually testing speci- 
mens whose smallest areas varied in length, to determine what 
must be the least length of that part of the specimen whose 
cross-section area is smallest, in order that the tensile strength 
may not be greater than with a long specimen. The conclusion 
reached was, that no test-piece should be less than one-half inch 
in diameter, and that the length should never be less than four 
diameters ; while a length of five or six diameters is necessary 
with soft and ductile metal in order to insure correct results.. 
The following results of testing steel are given in Mr. Kent's 
book, as confirming the same rule in the case of steel. The 
tests were made upon Bessemer steel by Col. Wilmot at the 
Woolwich arsenal. 



TESTS OF COMMANDED E-EARDSLEE. 



40 T 



By groove form 
By cylinder . . 



Tensile Strength. 



Highest 

Lowest 

Average 

Highest 

Lowest 

Average 



Pounds per 
Square Inch. 



162974 
136490 

153677 
123165 

103255 
I 14460 



3°. Commander Beardslee also noticed that rods of certain 
diameters of the same kind of iron bore less in proportion than 
rods of other diameters ; and, after searching carefully for the 
reason, he found it to lie in the proportion between the diam- 
eter of the rod and the size of the pile from which it is 
rolled. The following examples are given : — 

ij-in. diameter, 6.62% of pile, 56543 lbs. per sq. in. tensile strength. 



'i 


8.18% 


it 


56478 " 


u 


a 


a 


If 


9.90% 


a 


54277 " 


a 


a 


a 


I* 


H.78% 


a 


5355° " 


(( 


if 


a 


«# 


7.68% 


(( 


56344 '■' 


tt 


tt 


tc 


If 


8.90% 


n 


55018 " 


tt 


tt 


tt 


4 


10.22% 


a 


54034 " 


tt 


tc 


tt 


2 


H.63% 


(( 


51848 " 


tt 


tt 


tt 



He therefore claims, that, in any set of tests of round iron, 
it is necessary to give the diameter of the rod tested, and not 
merely the breaking-strength per square inch. 

4 . He gives evidence to show, that if a bar is under-heated, 
it will have an unduly high tenacity and elastic limit ; and that 
if it is over-heated, the reverse will be the case. 



402 



APPLIED MECHANICS 



5°. The discovery was made independently by Commander 
Beardslee and Professor Thurston, that wrought-iron, after 
having been subjected to its ultimate tensile strength without 
breaking it, would, if relieved of its load and allowed to rest, 
have its breaking-strength and its limit of elasticity increased. 

His experiments show that the increase is in irons of a 
fibrous and ductile nature, rather than in brittle and steely 
ones ; hence the latter class would be but little benefited by 
the action of this law. 

The most characteristic table regarding this matter is the 
following : — 

EFFECT OF EIGHTEEN HOURS' REST ON IRONS OF WIDELY DIFFER 
ENT CHARACTERS. 





Ultimate 


Strength 


— ' 1 

1 

1 




per Square Inch. 










Remarks. 








First 


Second 






Strain. 


Strain. 




Boiler iron . . . 


48600 


5 6 5°° 


Not broken. 


tt tt 


49800 


57000 


Broken -\ 


u tt 


49800 


58000 


Broken I Average gain, 


tt tt 


48100 


54400 


Broken | 15.8%. 


tt if 


48150 


5555° 


Broken J 


Contract chain iron, 


50200 


54000 


Broken ^ 

Not broken 1 Average 


tt tt « 


5° 2 5° 


53200 


tt a tt 


50700 


553°° 


Not broken f- gain, 
Not broken J 6.4%. 


tt u tt 


49600 


52900 


tt a ft 


51200 


52800 


Not broken J 


Iron K . . . . 


58800 


64500 


Broken -\ 

Broken L Avera S e § ain > 

Broken J 94 %* 

f 
) 


it tt 


59000 


65800 


1 tt ft 

i 


56400 


60600 



CHAIN CABLE. 403 



§ 233. Chain Cable. — The most thorough set of tests of the 
strength of chain cable is that made by Commander Beardslee 
for the United-States government, an account of which may be 
found either in the report already referred to, or in the abridg- 
ment by William Kent. 

In this report are to be found a number of conclusions, 
some of which are as follows : — 

i°. That cables made of studded links (i.e., links with a 
cast-iron stud, to keep the sides apart) are weaker than open- 
link cables. 

2 . That the welding of the links is a source of weakness ; 
the amount of loss of strength from this cause being a very 
uncertain quantity, depending partly on the suitability of the 
iron for welding, and partly on the skill of the chain-welder. 

3 . That an iron which has a high tensile strength does not 
necessarily make a good iron for cables. Of the irons tested, 
those that made the strongest cables were irons with about 
5 1000 lbs. tensile strength. 

4 . The greatest strength possible to realize in a cable per 
square inch of the bar from which it is made being 200 per 
cent of that of the bar-iron from which it was made, the cables 
tested varied from 155 to 185 per cent of that of the bar- 
iron. 

5 . The Admiralty rule for proving chain cables, by which 
they are subjected to a load in excess of their elastic limit, 
is objected to, as liable to injure the cable : and the report 
suggests, in its place, a lower set of proving-strengths, as given 
in the following table ; the Admiralty proving-strengths being 
ilco given in the table. 

In these recommendations, account is taken of the different 
Drcportion of strength of different size bars as they come from 
th.2 rolls, also no proving-stress is recommended greater than 
50 per cent of the strength of the weakest link, and 45.5 per 
cent cf the strongest ; v/hereas in the Admiralty tests, 66.2 



404 



APPLIED MECHANICS. 



per cent of the strength of the weakest, and 60.3 per cent of 
the strongest, is sometimes used. 

For the details of this investigation, see the report, Execu- 
tive Document No. 98, 45th Congress, second session, or the 
abridgment already referred to. 



Diameter of 

Iron, 

in inches. 


Recommended 


Admiralty 


Diameter of 

Iron, 

in inches. 


Recommended 


Admiralty 


Proving-Strains. 


Proving-Strains. 


Proving-Strains. 


Proving-Strains. 


2 


121737 


161280 


I* 


66138 


83317 


itf 


1 14806 


I5I357 


If 


60920 


76230 


J-8 


I08058 


I4I750 


ifV 


55903 


69457 


lit 


IOI499 


132457 


I* 


51084 


630OO 


If 


95128 


I23480 


I* 


46468 


56857 


iH 


88947 


II4817 


ll 


42053 


51030 


if 


82956 


106470 


IlV 


37820 


455*7 


iA 


77159 


98437 


I 


33 8 40 


40320 


i* 


71550 


90720 









While steel long ago displaced wrought-iron for boiler-plate, 
and while steel I beams, channel-bars, angle-irons, and other 
shapes, as well as eye-bars, have, of late years, displaced 
wrought-iron to a very great extent, nevertheless wrought-iron 
is still very extensively used, and for a great variety of struc- 
tural purposes. 

For wrought-iron to be used in construction, ductility, 
homogeneity, and often weldability are the great desiderata, 
together with as large a tensile strength as is consistent with 
these. As to the requirements made by different engineers for 
wrought-iron for structural purposes, the minimum tensile 
strength called for varies from about 46000 to about 50000 
pounds per square inch, with ultimate elongations varying from 
15$ to 30$ in 8 inches, according to the purpose for which it is 
wanted. It is also very common, when good iron is wanted, 



CHAIN CABLE. 



405 



to insist that it shall not be made of scrap. The following 
tables of tensile tests of wrought iron of various kinds will 
show what results can be obtained. 







N'orway 


Iron. 






Burden's 


Best. 






rt'.S 


c 


c 

4> 






is 


e 


c 

V 






. 










. 




«m O 








IS" 

3fc 


0^ 

c « 

a 




1 <n 


3 « 


fsr 

3s 


J a 


0) u 


w V 


i a 


y a 


zi 


3 z * 


> J3 


i a 


y a. 


« ?! 


3 - n 




•5 «5 




3° 


3 rt e 

•a — * 


ll 




ll 


3 JJ 


•a -2 


5 


S 


3 


<* 


S C 




S 


5 


PJ 


s 


•75 


48390 


23620 


62.6 


30090000 


76 


53566 


27554 


57-6 


29175000 


•75 


46340 


21160 


62.7 


30780000 


75 


50023 


26030 


49-8 


30643000 


•75 


48280 


28030 


62.6 


29020000 


76 


47724 


25350 


47-6 


30310000 


•77 


45160 


20400 


68.8 


27388000 


77 


46772 


24700 


45-2 


28347000 


•75 


46063 


19240 


68.6 


27666000 


77 


46600 


22550 


46.2 


29528000 


•77 


44490 


20510 


6 7 -5 


28452000 


77 


47395 


22550 


46.2 


28347000 


•74 


43233 


22079 


70.5 


29026000 


77 


47963 


22695 


48.6 


29475000 


•75 


43470 


19400 


75-5 


26700000 


77 


47860 


26948 


46.4 


26948000 


•73 


38950 


22030 


72.3 


30140000 


77 


47500 


26927 


42.3 


28435000 


•74 


43240 


21970 


75.2 


27726000 


76 


47610 


23036 


53 -t 


29551000 


•74 


44564 


21970 


7? -8 


28663000 


77 


49238 


22725 


49.2 


27470000 


•74 


43860 


19658 


75.0 


18000000 


76 


50037 


27700 


53-6 


29251000 


1 .00 


41620 


15560 


70.3 


27205000 


76 


48538 


27224 


48.8 


29355000 


•75 


42215 




68.6 


20292000 


76 


50060 


23201 


53.o 


31028000 


•75 


42033 


19239 


62.4 


29729000 


76 


49143 


23240 


50.4 


30438000 


.76 


41574 


14328 


69-5 


27450000 


76 


48655 


23414 


49.6 


30062000 


.76 


4>574 


16531 


68.7 


29098000 


76 


47220 


22880 


53-4 


29969000 


•75 


42666 


19240 


59-o 


31785000 


76 


47090 


23020 


54-i 


33657000 


•75 


4i875 


16978 


70.1 


30487000 


76 


49690 


27480 


53-3 


29614000 


•75 


43396 


19112 


59-3 


28000000 


76 


47430 


22950 


51.8 


29443000 


•74 


39210 


15216 


73.2 


30294000 


76 


47950 


23000 


57-o 


29504000 


•74 


40343 


12603 


70.5 


28810000 


76 


49381 


22892 


45-8 


28779000 


•74 


39896 


15187 


69.7 


31 153000 


77 


49411 


18420 


46.6 


30112000 


•74 


39156 


16123 


76.4 


29807000 


7 6 


49660 


23186 


51-3 


30160000 


•74 


41030 


17490 


69.8 


29310000 


76 


48055 


20940 


56.7 


28809000 


•75 


41180 


18000 


7*-5 


31073000 


77 


49026 


22578 


40.5 


27292000 


•74 


42320 


T9660 


68.0 


30834000 


76 


47220 


23060 


51-2 


33710000 


•74 


43913 


19833 


69.8 


26970000 


76 


50149 


20940 


4i-5 


27450000 


•74 


42102 


I9L8I 


78.3 


29127000 


75 


48553 


23767 


74-3 


31124000 


•74 


39698 


17638 


7o-5 


30023000 


75 


49350 


21503 


66.5 


31793000 


•73 


43187 


17846 


68.6 


28553000 


76 


50083 


20940 


5i-4 


29097000 


•73 


40669 


17820 


73-8 


30159000 


76 


47019 


23146 


51-4 


29978000 


•73 


39348 


16593 


69-3 


29518000 


76 


47504 


20942 


53-2 


28527000 


•73 


39671 


12987 


78.1 


28861000 


76 


47747 


20942 


49-5 


29874000 


•75 


39951 


16886 


77.2 


30020000 


76 


50927 


23453 


46.9 


28350000 


•74 


41093 


16400 


74.1 


28634000 


75 


5126a 


21182 


5i-9 


32551000 


•74 


40192 


14053 


76.3 


28627000 


75 


50930 


23770 


53-7 


29293000 


•73 


44470 


16844 


73-4 


31114000 


76 


50083 


23146 


45- a 


29097000 


•74 


4:940 


17523 


.78.5 


29373000 


75 


48168 


23767 


55-6 


29879000 


•74 


4253a 


16449 


70.5 


30410000 


76 


49500 


26500 


55-0 


3 1 600000 












75 


48400 


27200 


46.2 


28700000 












75 


47600 


27200 


50.1 


29300000 












77 


47200 


23600 


56.1 


27800000 












76 


46700 


24200 


41.8 


29700000 












77 


45600 


23600 


54-5 


28200000 



406 



APPLIED MECHANICS. 



Refined Iron. 


Wrought-iron Wire. 




rt.S 
O . 


c 


c 

V 








Is 

. 


a 


a 

V 

c u 
"" u 

C U 

0. 


CO*-' 


V m 


J 0- 

3 « 


"1st 
3 $3 


c V 

.2 a 


"- 1 s>. 


Kind of Wire. 


v <p 


►J cr 
S u 

B& 


3fc 


S.8 


0* 


a 


rt 


3'S 




VJZ 


a 


"u rt 


2£« 


8g 


■3 U5 


to SS 


3 « 


3 rt 

•§5 




rt »2 


'§! 


to 55 


3 « 
4)<« 


3.t! 3 

•a « a* 
o\scr> 


s 


2 


2~ 


Pi 


S 




P 


s 


E~ 


Pi 


3 




56270 
53450 


28293 
28990 


33.9 

22.0 


28618000 


Annealed wire 


• 135 
.136 


70400 
61500 


43800 


63.I 
75-0 






ID 
11 


26997000 


Annealed wire 






76 


55880 


29758 


33-5 


277 1 1000 


Annealed wire 


.135 


61100 


39800 


49.4 


23000000 




77 


53850 


29370 


33-3 


28718000 


Annealed wire 


.136 


59500 


39200 


71.2 


2550000a 




77 


52770 


33722 


14.8 


27355000 


Annealed wire 


•135 


45100 


35800 


76.8 


23500000 




74 


52770 


28829 


33-5 


29273000 


Annealed wire 


.136 


59800 


34000 


72.7 






77 


51320 


29294 


22 6 


28082000 


Annealed wire 


• 155 


62400 




57.5 






77 


537'/8 


27138 


25-4 


28659000 


Common wire 


.no 


90900 


64000 


52.3 


30300000 




74 


48882 


28822 


13 8 


28137000 


Common wire 


.109 


103000 




64.4 


27500000 




75 


49240 


28190 


14.0 


27520000 


Common wire 


.no 


104000 


60c 00 


5I.O 


22900000 




75 


50190 


30590 


17.8 


26237000 


Common wire 


•"3 


93700 


68200 


60.5 


25200000 




77 


5'4°o 


29256 


22.4 


25680000 


Common wire 


.080 


I 13000 


45800 


41.9 


27000000 




75 


47495 


30387 


12.2 


27613000 


Common wire 


.080 


I I 3000 


56700 


5I.O 


26500000 




75 


48352 


30574 


17-3 


27177000 


Common wire 


.079 


112000 


54300 


53-3 


26600000 




76 


47I5I 


25982 


75-4 


21628000 


Common wire 


.079 


120000 


73600 


28.1 


26100000 




77 


50351 


35720 


25-3 


27477000 


Common wire 


•079 


109000 


54300 


40.4 


26400000 




75 


48202 


28521 


14.7 


27888000 


Common wire 


.080 


98300 




43-8 


27100000 




75 


50703 


30558 


13.0 


23713000 


Annealed wire 


.081 


99600 




61.9 


26600000 




75 


49223 


30517 


15-2 


27126000 


Annealed wire 


.082 


93500 


50400 


64-3 






75 


49120 


29000 


17.8 


28290000 


Annealed wire 


.082 


86300 


50400 


68.5 


27100000 




75 
75 
76 


47060 
47 8 30 
5x300 


31700 
29400 
26000 


15-4 
17.8 
29.1 




Annealed wire 


.082 


89900 
97100 
93500 


54000 
57600 
39600 


51-7 
55-o 
65-7 


24900000 




29290000 
30100000 


Annealed wire 
Annealed wire 


.082 




.082 


26100000 




76 


52400 


35000 


29.1 


25400000 


Annealed wire 


.082 


71000 


50400 


67.1 


27000000 




75 


534oo 


29000 


24.9 


28200000 


Common wire 


.167 


57200 


45100 


65.6 






76 
75 
76 
76 

77 


52100 
54100 
51500 
52500 
77300 


26000 


29.1 
24.9 
24.6 
22 .3 
26.5 




Annealed wire 


.081 


95900 
93500 
67400 
61500 


55300 
43200 
40100 
47300 


60.4 
75-o 
569 
52.8 






29000 
26500 
24200 
34400 


27700000 
33100000 


Annealed wire 

Common wire 

Common wire 

Piano wire, 


.082 






.163 
.163 










26800000 






75 
75 


53100 
52900 


34000 
31700 


24.9 
27.2 


27200000 
26100000 


No. 13 
Piano wire, 


.031 


345000 






29500000 










76 
>oX 


51600 


28700 


22.3 


26000000 


No. 23 


.048 


262500 






29300000 


4-c 










01 


40700 
53100 




14.4 
24.6 
















76 


28700 
















76 


52200 


33100 


26.9 


31000000 














75 


50100 


31700 


22.6 


28700000 














76 


49400 


26500 


26.9 












: 


I 


02 


50300 


31800 


16.9 


27200000 












I 


01 


47000 


32500 


20.6 


27700000 












I 


01 


50400 


30000 


25.8 


28300000 












I 


02 


49600 


31800 


23-9 


26500000 












I 


01 


50200 


30000 


32.5 


26800000 












I 


02 


50500 


29400 


30.6 


26200000 












I 


01 


51400 


30000 


29.2 


28300000 












I 


02 


50400 




20.4 


28200000 












i 


02 


50200 


31800 


1 5- 1 


27200000 












I 


01 


48100 


30000 


32.5 


27700000 












I 


01 
77 
74 
74 
76 
76 


50600 
48600 
53900 
54000 
495oo 
53500 


30000 
25800 
27900 
25600 
26400 
30900 


27.5 
52.6 
38.6 
18.0 
41.8 
33-4 


25800000 
28000000 
29700000 
29700000 














27600000 













TENSILE TESTS OF WROUGHT IRON. 



407 



In Heft IV (1890) of the Mitt. d. Materialpriifungsanstalt 
in Zurich is an account of a set of tensile tests of wrought-iron 
and mild-steel angles, tees, and channels. The following is a 
summary of his results for wrought-iron shapes : — 

ANGLE-IRONS. 









■a 

rt 






c 










3* -a 


■-*wg 

■9Sj 


,i^^3 




Modulus of 


u 


Dimensions, 
Inches. 


u 

a 


6^2 




Is. 


Elasticity, 
Pounds per 


.0 




%i 


X 3 3 


S33 


2§3 


3« 


Square Inch. 


3 




V cti 


rt ° cr 


rt o* 










£* 


gfcc/) 


w -^ 


^a.w 


eO 








Lbs. 












2 


2.76 X 2.76 X 0.31 


16.53 


49910 


25020 


37680 


9-5 


23824000 


4 


2.76 X 2.76 X 0.51 


27.62 


49060 


20190 


32560 


ii. 7 


28070000 


6 


3-54 X 3.54 X 0.35 


26.21 


50620 


25310 


34130 


15.8 


28269000 


8 


3.15 X 3-54 X0.55 


38.51 


5 1 190 


25310 


32000 


16.4 


27786000 


10 


4.13 X 4-13 X 0.47 


35-48 


49200 


28010 


33130 


12.0 


28537000 


12 


4.13 X 4-J3 X 0.67 


55-24 


46070 


22750 


32280 


10. 2 


28554000 


M 


5.12 X 5- 1 * X 0.67 


62.09 


47780 


22610 


30430 


12.0 


27985000 


16 


5.12 X 5«i2 X 0.87 


102.61 


48490 




31 140 


".3 







TEE-IRONS. 



3 


3". 60 X 3" -35 


22.88 


52470 


25880 


37680 


14.2 


27615000 


4 


" " 


" 


49200 


23610 


34700 


15.5 


27672000 


7 


3";?4 X 3 / ' l ;94 


31.75 


51760 


21610 


38820 


11. 7 


27857000 


8 




54040 


18630 


354io 


21.3 


27743000 


9 


5 ".qo X 3 "-94 


46.91 


53610 


23600 


36970 


19.0 


27402000 


10 






52900 


22890 


359 8 ° 


US 


28255000 



CHANNEL-IRONS. 



X 


4.13 X 2.56 


28.43 


50630 


23329 


35120 


15-9 


27544000 


2 


11 11 




49200 


24170 


33700 


12.7 


27885000 


4 


4.13 X 2.64 


3i-65 


543 2 ° 


23040 


36690 


20.6 


27772000 


S 


6.93 X 2.83 


48.89 


51760 


24460 


35550 


19.9 


27658000 


6 


I. 11 


" 


54610 


23320 


34270 


17-5 


27999000 


7 


6.93 X 2.91 


54-43 


51900 


19620 


35690 


20.4 


27487000 


8 




52900 


24320 


30860 


14.5 


29663000 


9 


8.46 X 3-35 


85.68 


52050 


22330 


3456o 


20.9 


27701000 


10 


" 




53470 


24170 


36260 


17.0 


28710000 


12 


8.46 X 3-5o 


92.33 


52760 


22040 


34840 


11. 9 


28568000 



408 



APPLIED MECHANICS. 



TENSILE TESTS MADE SUBSEQUENTLY AT THE WATERTOWN 

ARSENAL. 

, Here will next be given, in tabulated form, the results of a 
number of tensile tests made on the government machine at the 
Watertown Arsenal. 

The following tables of results on rolled bars, from the Elmira 
Rollifig-Mill Company (mark L) and from the Passaic Rolling- 
Mills (mark S), are given in Executive Document 12, 47th Con- 
gress, 1st session, and in Executive Document I, A!jth Congress, 
2d session. 

SINGLE REFINED BARS. 



a 

d 

n 

§ 


.5 

<J 

.1 1 

°" 
CO 


Elastic Limit, in 
lbs., per Square 
Inch. 


Ultimate Strength, 
in lbs., per 
Square Inch. 


if 

.5 
3 


"8 

1 ^ 

2 8 


Appearance of 
Fracture. 


Modulus of Elas- 
ticity at Load of 
20000 Lbs. per 
Square Inch. 




a 
. 


3^ 

u a 


L 1 


3.06 


28500 


52710 


18.4 


33-3 


95 


5 


2698)450 




L 2 


3.06 


29500 


53630 


16.4 


36.0 


92 


8 


27826036 




L 3 


3.06 


290OO 


52090 


21.4 


34-6 


95 


5 


28419182 




L 4 


3.06 


290OO 


5 '440 


15.O 


20.3 


90 


10 


3088803O 




L 5 


6.46 


27500 


50500 


H-5 


27.6 


95 


5 


27826036 




L 6 


6.40 


27500 


50530 


*7-3 


22.3 


70 


30 


27 1 18644 




L 7 


6-39 


270OO 


50200 


18.0 


22.5 


95 


5 


27444253 




L 8 


3-24 


- 


51667 


22.0 


36.0 


70 


30 


28318584 


Round. 


L 9 


3.20 


- 


50844 


16.3 


22.0 


l S 


85 


27972027 


<< 


L 10 


3.20 


- 


53062 


21.0 


40.0 


95 


' 5 


28l 19507 


« 


S 11 


3.08 


28500 


48640 


*3-3 


24-3 


100 


Slightly 


27586206 




S 12 


3.08 


280OO 


50390 


16.9 


35-i 


100 





27586206 




s 13 


3-°5 


28500 


47050 


9.0 


22.0 


100 





27874564 




S 15 


6.40 


260OO 


49700 


17.1 


19.2 


85 


15 


. 29906542 




S 16 


6.40 


240OO 


49280 


l S-7 


17.7 


85 


J 5 


26490066 




S 17 


6.41 


24500 


48740 


14-3 


17-3 


80 


20 


28l 19507 




S 18 


3-J7 


24600 


49680 


19-5 


32.0 


100 


Slightly 


27972027 


Round. 


S 19 


3-17 


25870 


49338 


18.3 


38.0 


100 





29357798 


u 


S 20 


3-»7 


24920 


48864 


18.4 


37-o 


100 


Cinder 
at centre 


27729636 


" 



DOUBLE REFINED BARS. 



409 



DOUBLE REFINED BARS. 



[ 

j 

£ 
03 
8 


8 J 

2 1 


Elastic Limit, in 
lbs., per Square 
Inch. 


Ultimate Strength, 
in lbs., per 
Square Inch. 



00 

J* 

J .9 

w 


.2 ^ 
2 8 

r 


Appearance of 
Fracture. 


Modulus of Elas- 
ticity at Load of 
20000 lbs. per 
Square Inch. 




3 
1 ^ 




L 201 


3.06 


2900O 


53560 


x 5-3 


37-9 


100 





2763385! 




L 202 


3-03 


30OOO 


52650 


16.2 


20.6 


85 


15 


34042 553 




L 203 


3.06 


32500 


53500 


16.5 


27.5 


100 





28169014 




L 204 


3.06 


325OO 


54480 15.4 


24.8 


100 





29090909 




L 205 


6.33 


270OO 


51230 17.8 


24.2 


80 


20 


28119507 




L 206 


6.34 


2750O 


50500 I 17.6 


21. 1 


100 


Slightly 


29629629 




L 207 


6.34 


270OO 


5103O 21.4 


3i-9 


100 





278260S6 




L 208 


3.20 


- 


5OI56 ;22.7 


43° 


100 


Cup- 
shaped 


2C02IOI 5 


Round. 


L 209 


3.20 


- 


49937 22.6 


45-o 


100 


" 


2862254O 


" 


L 210 


3.20 


- 


50188 19.9 


43-o 


100 


" 


289855O/ 


a 


S 211 


3-05 


2950O 


51150 22.0 


3 J -5 


100 





3298969O 




S 212 


3-05 


28500 


51 1 10 22.0 


36.1 


100 





25559^5 




S 213 


3-" 


29500 


51860 


22.5 


39-2 


100 





26446280 




S 215 


6.31 


2750O 


50980 


19.1 


23.6 


95 


5 


29357798 




S 216 


6.38 


27000 


50770 


20.7 


29.6 


100 





2826855I 




S 217 


Z-33 


27000 


5!340 


19-3 


35*2 


100 





2807OI75 




S 218 


3-i7 


2461O 


50631 


20.4 


41.0 


100 





2862254O 


Round. 


S 219 


3-17 


— 


50915 


25-5 


44.0 


100 


Cup- 
shaped 


2826855I 


» 


S 220 


3-17 


- 


50205 


23-7 


44.0 


100 


«< 


2807OI75 


« 



The moduli of elasticity had not been computed in the 
report, but have been computed in these tables from the elon- 
gations under a load of 20000 lbs. per square inch in each case, 
as recorded in the details of the tests. 

In these reports are also to be found tensile tests of iron 
from other companies, as the Detroit Bridge Company, the 
Phoenix Company, the Pencoyd Company, etc. Some of these 



4io 



APPLIED MECHANICS. 



tests were made to determine the effect of rest upon the bar 
after it had been strained to its ultimate strength, also to 
determine the strength after annealing. The following table 
shows these latter results : 




< 



a «T H xi «r°° «iS en «5 *° «2 

CO3 C^3 jD 2X5 X> 3-0 3 

£2 ES Eo £ fa 



bo 

•<* 

bToo rt 

sis 

fa O 



bo So 
3 0>* cm 

2 - I&^S. 

£ s s£ st 

~ 2 Srt 2 * 

3 .-2 :S 3 £ 3 

Q fa fa fa 



vs. « 

2 2 2 

g ^ 
C C rt v 

£ 5 3 .S 
fa fa fa 



IBUiSuo caojj 

B3JV JO 
UOU0BJ1UO3 



to On 



"0 d 'qiSuai 

^BUiSuo vaojj 

uouBSuoig 

auojiu/} 



o 

O o 

«,£ 

</> 5> 
O 



gboctS 
S a . . 



2 vg £•& 8 <S S3- 



5-vg 



« o VO 
■»»■ «o ro 

vo m vo 






sis 5 . 



o o> 



55 fj 



•o e 



« « 



O 05 

c£ 



o ? 

u 



.3 u 



rt c 
o 



*£ 



■§"2 3 

CO 

C 3T3 • 
CO « 



• 1 



C he*- 1 

- ._ <o 

tO u 0) 



_<3 

rt--t 
c ^ 

he*- 1 

.-to 



c ^ ;s 

.3 V O. 

.5" « rt 

u o> « 

O Pi ffi 



C "O 

So* 



.3 ai «_, 
.&> S5rt^.„ 

i- nuns *< « 



rt rt *j 



« o a B 

rt o w~~ 
4j o oi rt 



c ~ 

.3 v 

* £ 

>- o> 

O Pi 



DOUBLE REFINED BARS. 



411 



Some tests were made to determine the values of the 
modulus of elasticity of the same iron for tension and for 
compression ; and these were found experimentally to be 
almost identical, as was to be expected. For these tests the 
student is referred to the reports themselves ; and only cer- 
tain tests on eye-bars of the Phoenix Company will be 
appended here. 



Arsenal 
Number. 


Outside 
Length, 
Inches. 


Gauged 
Length, 
Inches. 


Sectional 
Area, 
Sq. In. 


Ultimate 

Strength, 

Pounds 

per sq. in. 


Contraction 

of Area at 

Fracture, 

per cent. 


5" 
513 
518 


67.75 
67.80 
96.05 


50 

50 

75 


1.478 
I.94O 
5.I03 


40600 
39480 
46720 


16.8 

13.9 

8.1 



Quite a number of tests of the iron of different American 
companies are to be found in the " Report on the Progress of 
Work on the Cincinnati Southern Railway," by Thomas D. 
Lovett, Nov. 1, 1875. 

For these the student is referred to the report named. 



WROUGHT-IRON PLATE. 



The following table contains some tests of wrought-iron 
plate and bars made on the Government testing-machine at 
Watertown in 1883 and 1884 f° r the Supervising Architect at 
Washington, D.C. 



412 



APPLIED MECHANICS. 



S 3 



2* S 



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1. 

£* * : : : : 



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ills 

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8« 5.. .S « .2 






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ofe 



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(V « 



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SO + Tj- CO H CO S IO 



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OVO N S00 



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}q3p u'j 



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a c 

55 



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m O\oo co co h no ■* ro ooo mw>ooo on On w h n no 

M O ** <M N CO 00 NO 00 M OOl ONOO ON O\00 S ON O -*" - 



IO00 00 ONC 



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~l IT) IO IT) IO LT) 



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O M O *")NhM O N innNCXO OMNO'3-rO 

-<*-"* i- o>ioiomt o onoo 101010* oo no no soo s 






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ttmoo. 



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000000000000000000000000 OOOOOOO --0QQ000 

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•too tn s o s 



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tSWCNN(NlNPl(NlW(NMlNM«MNMNMMMMMMMl-l NvNCNIWNWN 



WROUGHT-IRON AND STEEL EYE-BARS. 

In the report of the Government tests for 1886 is given the 
following table of tensile tests of wrought-iron eye-bars. The 
wrought-iron ones were furnished by the General Manager of 
the Boston and Maine Railroad, and the steel ones by the 
Chief Engineer for the American Committee of the Statue of 
Liberty. 



COMPRESSIVE STRENGTH OF WROUGHT-IRON. 413 



WROUGHT-IRON EYE-BARS. 



' Dimensions. 




K 


Elongation. 


2 





i> to . 

CD o;^; 




Fracture. 




t-c 


& 




< 


Sfi 


k'o2 














u 

■56-f 




to 

10 

3> 

c 

A! 




II 


** _: 

MX 

££ 

■55 a 


<UrC 

$2 


£ * 




a 


■g«a 


O C (o 

.E c g 


c 
.2 




Appearance. 


►j 


S 


£ 


w 


H 


£ 



O 


O 


s 


S 


J 




1 Ins. 


Ins. 


Ins. 


Lbs. 


Lbs. 


% 


% 


% 


Lbs. 


Lbs. 






238.55 


5 .00 


1. 14 


22456 


45io5 


11. 7 


11. 6 


31.2 


28037000 


52763 


Stem 


Fibrous, traces 
of granulation. 


238.60 


5.00 


1. IS 


22610 


4454o 


9.4 


9.4 


29.6 


28125000 


50588 




Fibrous, 70%. 
Granular, 30%. 


238.57 


4.99 


1. 14 


21790 


43320 


7-8 


2.o 


26.4 


27950000 


48492 




Fibrous, 70%. 
Granular, 30%. 


238.64 


5 .00 


1. 16 


22410 


3955o 


5-i 


4-8 


9.8 


27355000 


54013 




Fibrous, 70%. 
Granular, 30%. 


1238.62 


6.05 1.44 


19750 


43260 


12.05 


1 2 .06 


24.8 


28800000 


43166 




Granular, 80%. 
Fibrous, 20%. 


238.62 

• 


6.05 


1.44 


22730 


42020 


6.5 


6.6 


19.2 


28301000 


41929 




Granular 5% at 
one edge, fi- 
brous for bal- 
ance of fracture. 



The gauged length of the bars was 180 inches. The moduli 
of elasticity computed between 5000 and 10,000 pounds per 
square inch. 

COMPRESSIVE STRENGTH OF WROUGHT-IRON. 

In regard to the compressive strength of wrought-iron, we 
may wish to study it with reference to — 

1°. The strength of wrought-iron columns; 

2°. The strength of wrought-iron beams ; 

3 . The effects of a crushing force upon small pieces not 
laterally supported ; 

4 . The effects of a crushing force upon small pieces laterally 
supported. 

1°. In this case it may be said that, by reference to the 
tests of wrought-iron bridge columns, the compressive strength 
per square inch of wrought-iron in masses of such sizes is given 
by the tests of the shorter lengths of such columns, i.e., by 
those that are short enough not to acquire, when the maximum 
load is just reached, a deflection sufficient to throw any appreci- 
ably greater stress per square inch on any part of the column in 
consequence of the eccentricity of the load due to the deflec- 



4*4 APPLIED MECHANICS. 

tion. The results thus obtained are naturally lower than we 
should expect to obtain in smaller masses. 

2 . In this case the evidence that there is goes to show that 
the compressive strength is the same as in the case of i°, and 
hence that it is less than the tensile strength. Indeed, the i?esults 
of tests of full-size beams show a modulus of rupture greater 
than the compressive strength, less than the tensile strength in I 
sections, and greater in circular sections; all this being what 
would naturally be expected. 

3°. If a small cylinder of ductile wrought-iron is tested with- 
out lateral support, and with flat ends, the friction of the ends 
against the platforms of the testing-machine comes in to interfere 
with the flow of the metal; and if, besides this, the ratio of length 
to diameter is so small as to prevent buckling, then the specimen 
will gradually flatten out, and it becomes impossible to find any 
maximum load, because the area of the central part is constantly 
increasing. 

4°. In this case the crushing strength per square inch that 
causes continuous flow, and also the maximum strength per 
square inch, is greater than that where the specimen has no 
lateral support. Hence follows, that in the case of wrought-iron 
rivets it is entirely safe to allow a bearing pressure in the neighbor- 
hood of 90,000 or 100,000 pounds per square inch, according to 
circumstances. 

§ 223. Wrought-iron Columns. — Until after about 1880 
there was but little experimental knowledge on this subject beyond 
the experiments of Hodgkinson, which have furnished the con- 
stants for Hodgkinson's, and also for Gordon's formula, as already 
given in § 208 and § 209. 

These formulae have been in very general use, and it is only 
of late years that we have been able to test their accuracy by 
tests on full-size wrought-iron columns. The disagreement 
of the formulae already referred to, with the results of the tests, 
has led to the proposal of a large number of similar formulae, 



WROUGHT-IRON COLUMNS. 415 

each, having its constants determined to suit a certain definite 
set of tests, and hence all these formulae thus proposed, which 
are, of course, empirical, and can only be applied with safety 
within the range of the cases experimented upon. 

A few of these will now be enumerated; and then will follow 
tables of the actual tests, which furnish the best means of deter- 
mining the strength of these columns ; and it would appear that 
it is these tables themselves which the engineer would wish to use 
in designing any structure. 

On the 15th of June, 1881, Mr. Clark, of the firm of Clark, 
Reeves & Co., presented to the American Society of Civil Engineers 
a report of a number of tests on full-size Phoenix columns, made 
for them at the Watertown Arsenal, together with a comparison 
of the actual breaking-weights with those which would have 
oeen obtained by using the common form of Gordon's formula 
for wrought-iron. The table is shown on page 416. 

The very considerable disagreement between the breaking- 
loads as calculated by Gordon's formula, and the actual break- 
ing-loads, led a number of people to propose empirical formulae 
of one form or another which should represent this set of 
tests, and also others which should represent some other tests 
on full-size bridge columns, which had been previously made 
in other places. 

Of these I shall only give those proposed by Mr. Theodore 

Cooper, which are as follows : — 

P f 

For square-ended columns . . . -j ~ 



1 + 



£-*>)' 



18000 



P f 

For pin-ended columns . . . . -r ~ 



■ + (;-»)' 

7 8000 



4t6 



APPLIED MECHANICS. 



Total 


Ultimate 

Strength, in 

lbs., by 

Gordon's 

Formula. 


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OcoM00MMr^Oi-ii-iOvOcor^coONNCO 

nn^rJ-NN r-^oo O»0\w O N N COVO CO CO 

cococococococococoro^TfTfTj-Tl-Tr'Tj-Tt 


till 


So 
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V 

6 




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cococococococococococococococO'^tJ-vo 


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n COO w 
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"3 


888888888888888888 

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0.190 
0.186 

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0.160 
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i 
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.n rT+nO cor^oOvOM tooo n Tf- n- rf 
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n r» vo ri- 
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,. 



WROUGHT-IRON COLUMNS. 



417 






■SSSSSSSni 




4 i8 



APPLIED MECHANICS. 



And he gives, for the values of/", 

For Phoenix columns f= 36000; 

" American Company's columns . • . ./= 30000; 
" box and open columns /= 31000. 

He deduces these values of / from some tests made in 1875 
by Mr. Bouscaren, combined with those, already referred to, 
made at the Watertown Arsenal. The box and open columns 
were made of channel-bars and latticing. The tables or dia- 
grams presented to justify the formulae proposed can be found 
in the "Transactions of the American Society of Civil Engineers " 
for 1882. 

Besides the above there will be given here tables of three 
sets of tests of full-size wrought-iron columns, viz. : 

i°. The series made at Watertown Arsenal, this being the 
most complete set of tests of full-size wrought-iron columns in 
existence. 

2 . A series of tests of Z-bar columns made by Mr. C, L. 
Strobel. 

3 . A few tests made at the Mass. Institute of Technology. 
Reference will also be made to the tests of Mr. G. Bouscaren, 
and to those made by Prof. Tetmajer, at the Materialprufungs- 
anstalt in Zurich. 

Graphical rep esentations, however, will first be given of the 



• 


• 




















— 
























• • 








































• 






• 


Vi 


j| 




«• 


■ 










*6000- 














•'\ 


I • < 


}\ 


# 


•Stf 


^jjr 


*•< 


p 














• 


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* 


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• 


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• 


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• 




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4 


i 


6 





8 





i( 


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is 


►0 


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u 





i. 


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400CO 
30000 



10000 



results of those tested at Watertown Arsenal, with the correspond- 
ing curves, representing (a) the formulae of Prof. Sondericker (see 



WROUGHT-IRON COLUMNS. 4 J 9 

page 417), and (b) that of Mr. Strobel (see page 418). These 
diagrams will be preceded by the corresponding formulae. 

A perusal of them will show that, for values of — less than 

P 
a certain quantity, which Mr. Strobel assumes as 90, and Prof. 
Sondericker as 80 for flat-ended, and 60 for pin-ended columns ; 

p 
the value of -— (i.e., the breaking-load divided by the area) is 

I P 

constant. For greater values of — the value of — decreases, and 

p A 

for this portion of the curve, Prof. Sondericker's formulae are as 

follows : 

For flat-ended Phoenix columns he recommends Cooper's 
formula. 

For lattice columns with pin-ends, reported in Exec. Doc. 
12, 47th Congress, 1st session, and Exec. Doc. 5, 48th Congress, 
1st session, he recommends the formula 

P 34000 

I+ Qr- 6 °) 

12000 
For the box and solid web columns reported in Exec. Doc. 5, 
48th Congress, 1st session, and Exec. Doc. 35, 49th Congress, 
1st session, taken together with Bouscaren's results on box and 
on American Bridge Company's columns, he recommends 

P 33000 



For flat- ends. 



A I 

1 + 

\P 



10000 

_ . , P 31003 

For pin-ends -7-= j- 7 r-^. 

A fl ' \ 2 

6000 



4 20 . APPLIED MECHANICS. 

In these formulae P = breaking-load in pounds, A= sectional 
area in square inches, /= length in inches, and p = least radius 
of gyration of section in inches. 

Moreover, the numerator in each of these formulae is the 

P I 

value of -j" corresponding to the case when - is less than 80 in 
A p 

flat-ended, and less than 60 in pin-ended columns. 

Instead of the above Mr. Strobel recommends for value of 

P v I - , -\ 

-r when — is less than 90, 35000 pounds per square inch, and, 

A p 

I 
for values of — greater than go, the formula 

p a l 

—r = 460OO — 1 2 5 — . 

A D p 

Moreover, if P'=safe load, in pounds, he recommends 

/ P' 

(a) For— < 90, ^- = 8000; 

/ P f I 

(b) For — > 90, -j- = 10600 — 30—. 

While Gordon's formula, or a modification of it, is still in 
use in many bridge specifications, quite a number of them have 
substituted the Strobel formula, or a modification of it. 

Wrought-iron columns subjected to eccentric load. 

All the formulae given thus far for the breaking or for the 
safe load on wrought-iron columns are only applicable when 
the load is so applied that its resultant acts along the axis of the 
column, and either the diagrams on pages 417 and 418, or the 
corresponding formulae, give us the breaking-strength per square 
inch, i.e., the number of pounds per square inch which, multiplied 



WROUGHT-IRON COLUMNS. 421 

by the area in square inches, gives the breaking-load of the column; 
the safe load per square inch being obtained by dividing the 
breaking-load per square inch by a suitable factor of safety. On 
the other hand, whenever the resultant of the load on the column 
does not act along the axis of the column, we must determine the 
fibre-stress due to the direct load, and to this add the greatest 
fibre stress due to the bending-moment, the sum of the two being 
the actual greatest fibre stress, and the column must be so pro- 
portioned that this greatest fibre stress shall not exceed the safe 
strength per square inch, as determined by dividing the breaking- 
strength per square inch by the proper factor of safety; and this 
proceeding should be followed whatever be the cause of the 
eccentric load — whether it be due to the beams supported by the 
column on one side being more heavily loaded than those on the 
other, whether it be due to the load transmitted from the columns 
above being eccentric, whether it be due to the mode of connection 
of the column to the other parts of the structure, whether it be 
due to poor fitting, or to any other cause. 

TESTS OF FULL-SIZE WROUGHT-IRON COLUMNS. 

The tests made at the Watertown Arsenal will next be given, 
together with cuts showing the form of the columns ; these being 
taken from the Tests of Metals for 1881, 1882, 1883, 1884, and 
1885. 

The following tables are taken from the volume for 1881.: 



422 



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424 



APPLIED MECHANICS. 



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WROUGHT-IRON CGLUMNS. 



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427 




428 



APPLIED MECHANICS. 




WROUGHT-IRON COLUMNS. 



429 



The next table taken from the volume for 1882" men- 
tioned above contains the results of some compressive tests 
of wrought-iron I-beams placed in the machine with the ends 
vertical and tested with flat-ends ; also of some tensile speci- 
mens cut off from two of them. 

TESTS OF I-BEAMS BY COMPRESSION. 





Length. 
In. 


Width 
of 

Flange. 
In. 


Thick- 
ness of 
Web. 
In. 


Total 
Depth. 

In. 


-t-j 

A 
bli 

L s. 


Sectional. 
Area. 

Sq. In. 


Ultimate 


Strength. 


Actual. 
Lbs. 


Per Sq. In 

Lbs. 


I 


57.06 


5-45 


O.64 


9.00 


228 


14.40 


545 TOO 


37854 


2 


155-45 


4.40 


O.4O 


IO.52 


443 


IO.26 


207000 


20170 


3 


191.90 


3-56 


0.40 


9.08 


365 


6.85 


85380 


12460 


4 


191.90 


3-59 


0.43 


9.09 


381 


7-15 


85200 


II916 


5 


119-85 


2.98 


O.28 


6. 11 


139 


4.18 


IOI200 


24210 


6 


180.33 


3.60 


O.42 


6.96 


303 


6.05 


84650 


13990 


7 


192.04 


3.58 


0.45 


7-94 


355 


6.65 


834OO 


12540 


8 


192.90 


3.60 


O.44 


7.98 


353 


6-59 


923OO 


I4OIO 


9 


215.88 


4.28 


O.40 


10.52 


' 56i 


9-30 


I49OOO 


I602O 


10 


264.08 


4.49 


O.48 


10.53 


747 


10.19 


II3IOO 


1 1 IOO 


11 


264 . 08 


4-43 


O.50 


10.51 


767 


10.46 


IO780O 


IO306 


12 


264 . 00 


4.90 


0-53 


15-15 


1085 


14.80 


1847OO 


12400 


13 


263.95 


4.84 


0.53 


14.74 


108 1 


14.74 


187OOO 


12686 



TESTS OF SPECIMENS FROM NOS. I AND 2 BY TENSION. 





Cut from 
Flange 
or Web. 


Width. 


Depth. 


Sectional 
Area. 


Ultimate 


Strength. 


Contrac- 
tion of 
Area. 


Actual. 


Per Sq. In. 






In. 


In. 


Sq. In. 


Lbs. 


Lbs. 


Per Cent. 


r 


Web. 


3.00 


O.65 


1-95 


103300 


52970 


IO 




Web. 


3.00 


O.50 


I-5I 


65400 


43340 


3-9 




Flange. 


4.00 


0.75 


3-OI 


146400 


48640 


19.6 


. 


Flange. 


4.OO 


O.76 


3.02 


147100 


48640 


15-9 


■ I 


Flange. 


3.00 


O.51 


1-53 


55400 


36210 


11. 1 


I 


Web. 


3.00 


O.40 


1. 19 


529OO 


44640 


16.5 



430 APPLIED MECHANICS. 

Next will be given the set of tests which is reported in the 
volumes for 1883 and 1884. 

The following is quoted from the first of the two : 



COMPRESSION TESTS OF WR0UGHT-IR0N COLUMNS, LATTICED, BOX, 
AND SOLID WEB. 

" This series of tests comprises seventy-four columns, forty 
of the number having been tested, the results of which are 
herewith presented. 

" The columns were made by the Detroit Bridge and Iron 
Company. 

" The styles of posts represented are those composed of — 

" Channel-bars with solid webs ; 

" Channel-bars and plates ; 

u Plates and angles ; 

" Channel-bars latticed, with straight and swelled sides ; 

" Channel-bars, latticed on one side, and with continuous 
plate on one side. 

" All the posts were tested with 3^-inch pins placed in the 
centre of gravity of cross-section ; except two posts of set N, 
which had the pins in the centre of gravity of the channel- 
bars. 

" This gave an eccentric loading for these columns, on ac- 
count of the continuous plate on one side of the channel- 
bars. 

" The pins were used in a vertical position, unless other- 
wise stated in the details of the tests. 

" In the testing-machine the posts occupied a horizontal 
position. 

" They were counterweighted at the middle. 

" Cast-iron bolsters for pin-seats were used between the ends 



WROUGHT-IRON COLUMNS. 43 I 

of the columns and the flat compression platforms of the test- 
ing-machine. 

" The sectional areas were obtained from the weights of the 
channel-bars, angles, and plates, which were weighed before 
any holes were punched, calling the sectional area, in square 
inches, one-tenth the weight in pounds per yard of the iron. 

" Compressions and sets were measured within the gauged 
length by a screw micrometer. 

"The gauged length covered the middle portion of the 
post, and was taken along the centre line of the upper chan- 
nel-bar or plate, always using a length shorter than the dis- 
tance between the eye-plates, to obtain gaugings unaffected by 
the concentration of the load at those points. 

" The deflections were measured at the middle of the post. 
The pointer, moving over the face of a dial, indicated the 
amount and direction of the deflection. 

" Loads were gradually applied, measuring the compressions 
and deflections after each increment ; returning at intervals to 
the initial load to determine the sets. 

" The maximum load the column was capable of sustaining 
was recorded as the ultimate strength, although, previous to 
reaching this load, considerable distortion may have been pro- 
duced. 

" Observations were made on the behavior of the posts 
after passing the maximum load, while the pressure was fall- 
ing, showing, in some cases, a tendency to deflect with a sudden 
spring, accompanied by serious loss of strength. 

" The slips of the eye-plates along the continuous plates 
and channel-bars during the tests were measured for certain 
posts in sets F, G, 77, and I. The measurements of slip were 
taken in a length of 10 inches or 20 inches, one end of the 
micrometer being secured to the eye-plate, and one end to the 
channel-bar. The readings include both the compression 
movement of the material and the slip of the plates. 



43 2 APPLIED MECHANICS. 

" Columns H, I, Z, and M were provided with pin-holes for 
placing the pins either parallel or perpendicular to the webs of 
the channel-bars. 

" After the ultimate strength had been determined with the 
pins in their first position, a supplementary test was made, if 
the condition of the column justified it, with the pins at right 
angles to their former position ; thus changing the moment of 
inertia of the cross-section, taken about the pin as an axis. 

" The experiments with columns N show how much strength 
is saved by employing pins in the centre of gravity of the cross- 
section. Where such was not the case, the columns showed 
the effect of the eccentric loading by deflections perpendicular 
to the axis of the pins, from the initial loads, which resulted in 
their early failure." 



WRO UGHT-IRON COL UAJXS. 



433 



TABULATION OF EXPERIMENTS ON WROUGHT-IRON COLUMNS 
WITH 3I-INCH PIN-ENDS. 













Length, 
Centre 

to 
Centre 

of Pins. 

In. 


Sec- 
tional 
Area. 

Sq. In. 


Ultimate 
Strength. 




No. of 
Test. 


Style of Column. 


Total, 
Lbs. 


Lbs. 

per 

Sq. In. 


Manner of Failure. 


752 


1 


Set 

1 * 


A. 

1 


f 



[ 


126.20 
120.07 
180.00 
180.00 
240.00 
240.10 

240.00 
240.00 
320.00 
320.10 

320.00 
320.00 

3I9-95 
320.00 


9-831 
10.199 

9-977 
9-977 
9-732 

9.762 

16.077 
16.281 
16.179 
16. 141 

17.898 
19.417 

16.168 
16.267 


297100 
320000 
251000 
210000 
188600 
158300 

425000 
367000 
318800 
283600 

474000 
491000 

453ooo 
454000 


30220 
31380 
25160 
21050 
19380 
16220 

26430 
22540 
19700 
17570 

26480 
25290 

28020 
27910 


Deflected perpendicular 
to axis of pins. 


757 
755 
756 
753 
754 

75i 




CD 


Sheared rivets in eye- 
plates. 

Deflected perpendicular 
to axis of pins. 
Do. do. 

Do. do. 

Do. do. 

Deflected perpendicular 


m 

1 

'Set',D, 

1 1 


1642 
1646 
1647 


ir j 
— iii — .. 


to axis of pins. 
Do. do. 

Do. do. 

Do. do, 




I 


\ T 1 


> 




*°53 
i654 


few 


J 

CO 

1. 


Deflected perpendicular 
to axis of pins. 
Do. do. 




1 " 

Se't'F. 






1645 

1650 


*j 




tf 

1.1 


r 

L 


Deflected parallel to axis 

of pins. 
Deflected perpendicular 
to axis of pins. 




"> 













434 



A PPLIE I) ME CHA A ICS. 



TABULATION OF EXPERIMENTS ON WROUGHT-IRON COLUMNS 
WITH 3I-INCH PIN-ENDS. 



No. of 
Test. 



165 1 
1652 



746 
747 
748 
749 
1648 
1649 

740 
74i 
739 
75o 
1643 
1644 



1640 
1641 
1634 
1635 



Style of Column. 



Set- G. 



"1 
_1 



=BT 



S. 1 



L 



Set 


H 




1 
8 


c 



:t 



set 1. 

(swelled.) 



j_4 



SetL, 



_. 1 _ 



Length, 
Centre 

to 
Centre 

of Pins. 

In. 



320.00 
320.10 

159.20 

159-27 
239.60 
239 . 60 
3i9-9o 

3*9-85 

i59 -9° 
i59-9o 
239.70 
239.70 
319.80 
319.92 



199.84 
200.00 
300.00 
300.00 



Sec- 
tional 
Area. 



Sq. In. 



20.954 
20.613 



7.628 
8.056 
7.621 
7.621 
7-7°5 
7-673 

7-645 

7.624 
7-517 
7-531 
7.691 
7.702 



11.944 
12.302 
12.148 
12.175 



Ultimate 
Strength. 



Total, 
Lbs. 



540000 
535000 



258700 
294700 
260000 
254600 
243600 
229200 

262500 
255650 
251000 
259000 
237200 
237000 



Lbs. 

per 

Sq. In. 



403000 
426500 
408000 
395000 



25770 
25950 



339io 
36580 
34120 
334io 
31610 
29870 

3434o 
3353o 
3339o 
3439o 
30840 
30770 



Manner of Failure. 



Deflected in diagonal 
direction. 

Sheared rivets in eye- 
plates. 



Deflected perpendicular 
to axis of pins. 
Do. do. 



Do. 



do. 
in diagonal 



Deflected 

direction. 
|Deflected parallel to ?xis 

of pins. 
Deflected in diagonal 

direction. 



Deflected perpendicular 
to axis of pins. 
Do. do. 

Deflected parallel to axis 

of pins. 
Deflected perpendicular 

to axis ol pins. 
Deflected parallel to axis 

of pins. 
Deflected in diagonal 

direction. 



3374o 
34670 
33630 
32440 



Deflected perpendicular 

to axis of pins. 
Deflected in diagonal 

direction. 
Deflected perpendicular 

to axis of pins. 
Do. do. 



WR0UGHT-1R0N COLUMNS. 



435 



TABULATION OF EXPERIMENTS ON WROUGHT-IRON COLUMNS 
WITH 3J-INCH PIN-ENDS. 















Length, 
Centre 

to 
Centre 

of Pins. 

In. 


Sec- 
tional 
Area. 

Sq. In. 


Ultimate 
Strength. 




No. of 
Test. 


Style of Column. 


Total, 
Lbs. 


Lbs. 

per 

Sq. In. 


Manner of Failure. 




SejtM. 












1638 
1639 
1636 




(swelled.) 

........ 




5 


199-25 
199.50 
300.20 


12.366 
12.659 
11.920 


385000 
405000 
391400 


3"3o 
31990 
32830 


Deflected perpendicular 
to axis of pins. 
Do. do. 

Deflected in diagonal 
direction. 

Do. do. 


«637 


-— 


1 

""I"" 

1 






3°o- 15 


11.932 


390700 


32740 




| ! 

SejtN. 












1630 


II 


£2, 


„- 


i 


300.00 
300.00 
300.00 
300.00 


17.622 
17.231 
i7-57o 
17.721 


461500 

485000 
306000 
307000 


26190 
28150 
17420 
17270 


Deflected perpendicular 


1631 
1632 
1633 


, °! 


. 1 


■"* 


to axis of pins. 
Do. do. 

Do. do. 

Do. do. 


3 lD 


a" 






II g; t 








I 1 

< 


1, 

















The remainder of the tests of this series of seventy-four 
columns is reported in the volume for 1884. 

The only portion of the description that it is worth while 
to quote is the following, as the tests were made in a similar 
way to what has been already described : 

" Sixteen posts were tested with flat ends ; eighteen were 
tested with 3j-inch pin-ends. 



436 



APPLIED MECHANICS. 



" The pins were placed in the centre of gravity of cross- 
section, except two posts of set K, which had the pins in the 
centre of gravity of the channel-bars, giving an eccentric bear- 
ing to these columns, on account of the continuous plate on 
one side of the channel-bars." 



TABULATION OF EXPERIMENTS ON WROUGHT-IRON COLUMNS 
WITH FLAT ENDS. 











Total 
Length. 

Ft. In. 


Sec- 
tional 
Area. 

Sq. In. 


Ultimate 
Strength. 




No. of 
Test. 


Style of Column. 


Total, 
Lbs. 


Per 

Sq. In., 

Lbs. 


Number of Failure. 




SetB. 
SetE. 


1 J 


-x" 


10 

10 

13 
13 

x 3 
13 
20 

20 

13 

13 

20 
20 


7.90 
7.90 

11.80 
11.80 

11. 9 
11.65 
763 

7.80 

"•75 
7.60 

7-63 


12.08 

17.01 
17.80 

15-74 
15.84 
15.68 
15.56 

21 .02 
21.46 
21 .20 

21.49 


383200 
372900 

594500 

633600 

517000 
555200 
517500 

536900 

708000 
709500 
700000 
729450 


31722 
33564 

3495o 
35595 

32846 
35050 
33003 
34505 

33682 
33061 
33019 
33943 




377 
378 


^1 


Buckling-plate D be- 
tween the riveting. . 
Buckling-plates. 




I I 




379 

380 


to 


Buckling - plates be' 

tween the riveting. 
Triple flexure. 


346 
347 
342 




Buckling-plates. 

Do. do. 
Deflecting upward. 


~1 

Set F. 

J 




I 
L 


344 
348 


*9tf' 


Buckling-plates. 
Buckling-plates. 


349 
34i 


1 «" ,T 

SetG. ^rJfc" § 
J^ 6.90 '-fc 


! 


Triple flexure. 
Deflecting upward. 
Deflecting downward. 


343 


^a" 



WRO UGH T-IRON COL UMNS. 



43 



TABULATION OF EXPERIMENTS ON WROUGHT-IRON COLUMNS 
WITH FLAT ENDS. 



No. of 
Test. 



339 
340 



337 
338 



Style of Column. 



Ka Plate. 



SetK. 



IT 



SetN. 



Y 16 plate.' 



Total 
Length. 

Ft. In. 



20 7.94 
20 7.94 



25 7-75 
25 7-88 



Sec- 
tional 
Area. 

Sq. In, 



12.64 
12.74 



16.99 
17.40 



Ultimate 
Strength. 



Total, 
Lbs. 



412900 
431400 



582400 
580000 



Per Sq. 
In., lbs. 



32666 
33862 



Manner of Failure-. 



Deflecting upward. 
Do. do. 



34279 Deflecting downward 

and sideways. 
33333 [Deflecting diagonally 
j channel B and lattic- 
ing on the concave 
1 side. 



TABULATION OF EXPERIMENTS ON WROUGHT-IRON COLUMNS 
WITH 3J-INCH PIN-ENDS. 



Ho. of 
Test. 



368 
367 
356 
357 

37i 
372 
370 
3 6 9 
354 
365 



Style of Column. 



Set B. 






%* 



Set C. 






Length, 
Centre See- 
to tional 
Centre 

of Pins.! 

Ft. In.! Sq. In. 



Area. 



15 0.1 

15 0.0 

20 0.0 

20 0.0 

9 «-9 

10 0.0 

15 0.0 

15 0.0 

20 0.0 

20 0.0 



11.42 
11.42 
11.42 
11. 31 

9.14 
10.07 
9.21 
9-44 
9.24 
9.36 



Ultimate 
Strength. 



Total, Per Sq. 
Lbs. In., lbs, 



379200 33205 
360200 32329 



342000 
330100 



29947 
29186 



Manner of Failure. 



Hor. deflection perpen- 
dic. to plane of pins. 
Hor. deflection. 



Do. 
Do. 



286100 31302 I Buckling - plates be- 



319200 
291500 

290000 ! 
267500 J 
279700 j 



31698 
31650 



tvveen rivets. 
Do. do. 



Hor. deflec. and buck- 
ling between rivets. 
30720 I Do. do. 

28950 Triple flexure. 

29879 Hor. deflection. 



438 



APPLIED MECHANICS. 



TABULATION OF EXPERIMENTS ON WROUGHT-IRON COLUMNS 
WITH 34-INCH PIN-ENDS. 



No. of 
Test. 



Style of Column. 



Length, 

Centre 

to 


Sec- 
tional 


Ultimate 
Strength. 






Centre 
of Pins. 


Area. 


Total, 
Lbs. 


Per 

Sq. In., 


Ft. In. 


Sq. In. 


Lbs. 


13 4-i3 


15-34 


475000 


30965 


13 4.00 


15.40 


485000 


3M94 


20 0.0 


17.77 


570000 


32077 


20 0.0 


17.22 


5554°o 


32253 


20 0.25 


12.48 


202700 


16242 


20 0.00 


10.84 


208200 


19207 


20 0.25 


12.65 


350000 


27668 


20 0.25 


12.76 


390400 


30596 



Manner of Failure. 



360 
361 



358 
359 



350 
351 
352 
353 



SetD 






:tE. 






p I ? 



4- 



fr%" 



SetK. 



W-rS-TrJ 
OS) '«> T 

si i§~ I 



Rfirn 



Deflecting upward in 
plane of pins. 

Hor. deflection perpen- 
dicular to plane of 
pins. 



Hor. deflection perpen- 
dicular to plane of 
pins. 

Do. do. 



Hor. deflection, con- 
cave on lattice side. 
Do. do. 



Do. 



do, 



30596 Hor. deflection perpen- 
dicular to plane of 
pins, convex on lat- 
tice side. 



Besides the above, there are four tests of lattice columns 
reported in Exec. Doc. 36, 49th Congress, 1st session, but as 
these columns were rather poorly constructed and form rather 
special cases they will not be quoted here. 

In determining the strength of a bridge column made of 
channel-bars and latticing, these results of tests on full-size 
columns furnish us the best data upon which to base our con- 
clusions. 



WROUGHT-IRON COLUMNS. 



439 



In the Trans. Am. Soc. C. E. for April, 1888, Mr. C. L. Stro- 
bel gives an account of his tests on wrought-iron Z-bar columns, 
from which the following is condensed, viz.: The Z-irons used 
in making the columns were 2JX3X2J inches in size, and & 
inch thick. 

Two columns were about 11 ft. long, two 15 ft., two 19 ft., 
three 22 ft., three 25 ft., and three 28 ft., a total of fifteen columns. 
The table of results follows: 



Length, 
Inches. 


Sectional 
Area, 

Sq. Ins. 


Ultimate 
Strength 
by Tests 
per Sq. In. 

Lbs. 


l_ 
p' 


Ultimate 
Strength, 

by 

Strobel's 

Formula 

per Sq. In. 

Lbs. 


I3ii 
131* 


9-435 
9.984 


36800 
34600 


64 
64 


— 


180 
180 


9.480 
9.280 


34600 
36600 


88 
88 


35000 
35000 


2 28f 


9.241 


33800 


112 


32200 


228f 


10.104 


33700 


112 


32200 


264 


9.286 


30700 


129 


29900 


264 
264 


9.286 
9.286 


29500 
30700 


129 
129 


29900 
29900 


300 
3OO 
300 


9.156 

9-45 6 
9.516 


28100 
28000 
28400 


146 
146 
146 


27750 
27750 
2 7750 


336 
336 

336 


9-375 
9-643 
9-375 


27700 
28000 
27600 


164 
164 
164 


25500 
25500 
25500 



The following table shows the results of compression tests 
made in the engineering laboratories of the Massachusetts In- 
stitute of Technology upon some wrought-iron pipe columns. 
They were tested with the ordinary cast-iron flange-coupling 
screwed on to the ends, bearing against the platforms of the 
testing-machine, which were adjustable, inasmuch as they were 
provided with spherical joints. 



440 



APPLIED MECHANICS. 



The tests of full-size wrought-iron columns made by Mr. G. 
Bouscaren, are given in the Report of the Progress of Work on 
the Cincinnati Southern Railway, by Thos. D. Lovett, Nov. i, 

1875. 

In Heft IV (1890) of the Mittheilungen der Materialpru- 
fungsanstalt in Zurich is given an account of a large number 
of tests of wrought-iron and steel columns of the following 
forms, viz.: i°. Angle-irons; 2° Tee iron; 3 . Channel- bars ; 
4*. Two angle-irons riveted together ; 5 . Four angle-irons 
riveted together; 6°. Two channel-bars riveted together; 
7°. Two tee irons riveted together ; also quite a number of tests 
of columns of some of these forms subjected to eccentric 
loads, the eccentricity of the load being, in some cases, as much 
as 8 cm. (3". 15). The columns tested were of a variety of 
lengths, the longest ones being 560 cm. (18.37) ^ eet l° n g- 

In Heft VIII (1896) of the same Mittheilungen is an ac- 
count of another set of tests of columns of the above-described 
forms. The results of these valuable tests will not be quoted 
here, but for them the reader is referred to the Mittheilungen 
themselves. 



a v 

U 


u 






•0 




•d 




- -a 




'•3 a 


V 

B 

S 
'Jo 


a 
5 

V 

U5 V 

3 <U 


SiS 


c 
So 


bin 
C 

u 

a 

3 

erj 


3 

8 

3 

a 

K 


M 

Cfl 

2 
H 


. 
►J a 

gen 

'Is. 


c"3 . 

.2 Z> >> 
"35 3 .- 

§2W 


-, or Ratio of 

Length to Ra 
us of Gyratio 


X 




O 


G 




O 


S 

Lbs. 


< 

Sq. In. 


S 


U 


In. 


In. 


In. 


In. 


In. 


In. 


Lbs. 






2 


2.06 


2-37 


7 


69H 


51 


30000 


1.08 


27800 


24300000 


88.8 


2 


2.04 


2-39 


7 


69H 


5i 


29800 


1 


22 


24500 


2 2 200000 


89.1 


4 


2.50 


2.89 


8 


93H 


86 


34500 


1 


65 


20900 


25200000 


98.1 


*i 


2.48 


2.88 


8 


93i 


86 


37000 


1 


68 


22000 


25900000 


98.4 


3 


3.06 


3-44 


H 


9311 


86 


455°° 


1 


94 


23500 


27700000 


81. 4 


3 


3.00 


3.48 


8* 


93§ 


86 


51000 


2 


01 


25300 


25100000 


80.5 


34 


3. bo 


4.00 


9k 


iosl 


100.5 


55000 


2 


39 


23000 


25200000 


78.2 


33 


3-59 


3-99 


9* 


105I 


100.5 


65000 


2 


39 


27200 


24600000 


78.5 


4 


4.07 


4-53 


9k 


"7t 7 b 


100.5 


80000 


3 


11 


25700 


25800000 


77.1 


4 


4.09 


4-5o 


9i 


"7i% 


100.5 


69000 


2.76 


25000 


24900000 


77-3 



§ 224. Transverse Strength of Wrought-iron. — 

Wrought-iron owes its extensive introduction into con- 
struction as much or more to the efforts of Sir William 
Fairbairn than to anyone else; and while he was furnishing 



TRANSVERSE STRENGTH OF WROUGHT-IRON. 44 l 

the means to Eaton Hodgkinson to make extensive experiments 
on cast-iron columns, and while he made experiments himself 
on cast-iron beams, which were in use at that time, he also 
carried on a large number of tests on beams built of wrought- 
iron, more especially those of tubular form, and those having 
an I or a T section, and made of pieces riveted together. In 
his book on the " Application of Cast and Wrought Iron to 
Building Purposes " he gives an account of a large number 
of these experiments, including those made for the purpose of 
designing the Britannia and Conway tubular bridges, a fuller 
account of which will be found in his book entitled " An Ac- 
count of the Construction of the Britannia and Conway Tubular 
Bridges." In the first-named treatise he urges very strongly 
the use of wrought-iron, instead of cast-iron, to bear a trans- 
verse load. 

Fairbairn tested a number of wrought-iron built-up beams, 
but they were of small dimensions and are hardly comparable 
with those used in practice. 

In the light of the tests made upon wrought-iron columns, 
it is evident that the compressive strength of wrought-iron is 
less than the tensile strength. Hence we should naturally ex- 
pect that the modulus of rupture would be, in all cases, greater 
than the compressive strength, and that it might or might not 
be greater than the tensile strength of the iron. Of course 
the modulus of rupture varies very much with the shape of the 
cross-section, for the same reasons as were explained in the 

paragraph 191, i.e., that the formula M = /— assumes Hooke's 

y 

law, " the stress is proportional to the strain," to hold, and that 
this is not true near the breaking-point. 

The value of the modulus of rupture is also influenced 
by the reduction in the rolls, and hence somewhat by the size 
of the beam. 

Small round or rectangular bars tested for transverse 
strength show a modulus of rupture very much in excess of 
the compressive strength per square inch of the iron, and ex- 
ceeding very considerably even the tensile strength. 

While a great many tests of such specimens have been 



44 2 APPLIED MECHANICS. 

made, none will be quoted here, but the last five tests of the 
table on page 542 show that for a wrought-iron having a ten- 
sile strength per square inch from 58700 to 60250 pounds, mo- 
duli of rupture were obtained from 80000 to 90000 pounds, 
as, the number of turns of these rotating shafts being com- 
paratively small, the breaking-loads were not far below the 
quiescent breaking loads. On the other hand the moduli of 
rupture of I beams and other shapes used in building have 
very much lower values, but for these, tests will be cited. 

As to experiments on large beams, we have : 

i°. Some tests made by Mr. William Sooy Smith and oy 
Col. Laidley at the Watertovvn Arsenal. 

2°. Some tests made in Holland on iron and steel beams, 
an account of which is given in the Proceedings of the Brit- 
ish Institute of Civil Engineers for 1886, vol. lxxxiv. p. 412 et 
seq. 

3 . Some tests made in the laboratory of Applied Mechan- 
ics of the Massachusetts Institute of Technology, on iron and 
steel I beams. 

4 . Tests made by the different iron companies upon beams 
of their own manufacture, and recorded in their respective 
hand-books. 

Mr. Smith's tests are recorded in Executive Document 23, 
46th Congress, second session. 

5 . In Heft IV (1890) of the Mittheilungen der Material- 
priifungsanstalt in Zurich will be found accounts of tests 
made by Prof. Tetmajer upon the transverse strength of 
I beams, of deck-beams, and of plate girders. 

The results of these tests will be given in the table on top 
of page 443- 

Specimens cut from the flanges, and also from the webs of 
the last seven of these beams, were tested for tension. In the 
case of those cut from the flanges, the tensile strength varied 



TRANSVERSE STRENGTH OF WROUGHT-IRON. 



443 



Depth. 
(Inches.) 


Moment of 

Inertia. 
(Inches)*. 


Span. 
(Inches.) 


Modulus of 

Rupture. 

(.Lbs. per Sq. In.) 

5 1 190 

56453 
62852 

56453 
53894 
51619 
53183 
53894 
52472 


Modulus of 

Elasticity. 

(Lbs. per Sq. In.) 


7.87 
7.87 
3-93 
5-9i 
7.87 
9-45 
11. 81 
J 3-39 
15-75 


52.04 
52.04 
4-13 
1785 
5 I -9S 
103.^4 


62.96 
62.96 
31-44 
47.28 
62.96 
75.60 
94.48 
107.12 
126.00 


27501500 
28937700 
28767101 
28212500 
28226700 
2 73735°° 
26463400 
27700600 
27679500 





from 50200 in the 1 5". 75 beam to 57300 pounds per square inch 
in the 3^.93 beam. On the other hand, in the case of the 
specimens cut from the web, the tensile strengths varied from 
44900 in the 1 i^.Si beam to 54400 pounds per square inch in the 
3".93 beam, the contraction of area per cent varying from 23.6 
to 32.1 per cent in the flanges, and from 12.5 to 15.9 per cent 
in the web. 

The results obtained with the deck-beams are as follows : 



Depth. 
(Inches.) 


Moment of 
Inertia. 
(Inches)*. 


Span. 
(Inches.) 


Modulus of 

Rupture. 

(Lbs. per Sq. In.) 


Modulus of 

Elasticity. 

(Lbs. per Sq. In.) 


4-93 
4.26 
3-52 
3 48 
2.36 
i-93 


19.88 
9-38 
5-33 
4.71 
1.30 
0.60 


70.86 
59.06 
47.24 
39-37 
3i-5o 
23.62 


56170 
48920 

553 2 ° 
54180 
52760 
58160 


251 12500 
25823500 
25596000 
26804700 
24202400 



Tensile tests of specimens cut from these deck-beams 
showed tensile strengths of from 47540 in the i".93 beam to 
54750 pounds per square inch in the 2". 36 beam, and contrac- 
tions of area of from 14. 1 per cent to 18.4 per cent. 

The results obtained with the plate girders are as follows, 
viz. : 



Depth of 

Web. 
(Inches.) 


Modulus of 

Rupture. 

(Lbs. per Sq. 

In.) 


Modulus of 

Elasticity. 

(Lbs. per Sq. 

In.) 


Depth of 

Web. 
(Inches.) 


Modulus of 

Rupture. 

(Lbs. per Sq. 

In.) 


Modulus of 

Elasticity. 

(Lbs. per Sq. 

In.) 


15-75 
J5-75 
19.69 
19.69 


51480 
53180 
51476 
52610 


26449200 
25539100 

24813900 
25695500 


23.62 
23.62 
27.56 
27.56 


52760 
48490 
47780 
46500 


26321200 ' 
26548700 
25667100 
26776300 



444 



APPLIED MECHANICS. 



The tensile strength of the material of the webs varied from 
29860 to 41240 pounds per square inch, while the contraction of 
area was only 0.4 per cent. The tensile strength of the material 
of the flange-plates was 51050 pounds per square inch, with a 
contraction of area of 17 per cent. The tensile strength of the 
angle-irons was 46357 pounds per square inch, with a contraction 
of area of 14 per cent. 

The following table gives the results that have been obtained 
in the tests that have been made upon wrought-iron I beams in 
the laboratory of Applied Mechanics of the Massachusetts Insti- 
tute of Technology. This table will give a fair idea of the strength 
and elasticity of such beams. 

TESTS OF WROUGHT- 1 RON BEAMS MADE IN THE LABORATORY OF APPLIED 
MECHANICS OF THE MASSACHUSETTS INSTITUTE OF TECHNOLOGY, 
ALL LOADED AT THE CENTER. 



No. 

of 

Test. 


i 
& 


Moment 

of 
Inertia. 


Span. 


Break- 
ing Load. 


Moduli 

of 

Rupture. 

Lbs. per 

Sq. In. 


Moduli of 

Elasticity. 

Lbs. per 

Sq. In. 


Remarks. 




Ins. 




Ft. 


Ins. 


Lbs. 








121 


6 


24.41 


12 





9500 


42386 


26679000 


From Phoenix Co. 


124 


7 


43-5° 


14 





14100 


48082 - 


2S457000 


< • it : > 


126 


5 


12.47 


12 





6450 


46624 


29549000 


ti «i ti 


209 


7 


44-93 


13 


8 


12200 


39670 


31057000 


t > t < < •• " 


211 


8 


67.32 


13 


8 


17000 


42000 


28532000 


it < « 1 < 


215 


9 


110.78 


14 


8 


23000 


41680 


27165000 


• 


227 


8 


61.20 


13 


6 


18300 


49640 


27397000 


From Belgium. 


230 


9 


86.41 


13 


8 


21300 


45 8 5° 


27365000 


c <• 


235 


9 


40.91 


13 


8 


13800 


49140 , 


27923000 


< c 1 • 


253 


7 


43-05 


14 


8 


11319 


40660 


28045000 


From Phoenix Co. 


256 


8 


66.56 


14 


8 


14547 


38460 


28187000 


« » <« it 


263 


9 


108.67 


14 


8 


19694 


36160 


27050000 


« < < 1 it 


291 


7 


45 -9 6 


14 


6 


10700 


36340 


26790000 


"•■ < « < 1 


292 


8 


66.39 


14 


6 


14300 


38200 


27380000 


«( t * < ' 


294 


9 


92.89 


14 


6 


19200 


41470 


27050000 


• 1 •. • t > 


338 


6 


25.92 


14 


7 


7200 


37800 


2 7860OOO 


N. J. Steel & Iron Co. 


341 


7 


46.73 


12 


11 


13600 


40600 


274IOOOO 


1 11 ti < < n 


345 


8 


71-25 


14 


7 


15400 


38400 


2694OOOO 


< t » : -i it 11 


379 


7 


48.84 


12 


10 


15500 


44300 


26170OOO 


1 > t < ! ■ » ' II 



STEEL. 445 



§ 225. Steel. — While steel is a malleable compound of iron, 
with less than 2 per cent of carbon and with other substances, 
the definition recommended by an international committee of 
metallurgists in 1876, and used to some extent in German and 
Scandanavian countries, is different from that in general use in 
English-speaking countries, and in France. 

The definition recommended by the international committee 
may be found in the Trans. Am. Inst. Min. Engrs. for October, 
1876, and is in the following: 

i°. That all malleable compounds of iron with its ordinary 
ingredients, which are aggregated from pasty masses, or from 
piles, or from any form of iron not in a fluid state, and which 
will not sensibly harden and temper, and which generally 
resemble what is called u wrought-iron," shall be called weld-iron 
(Schweisseisen). 

2 . That such compounds, when they will, from any cause, 
harden and temper, and which resemble what is now called 
puddled steel, shall be called weld-steel (Schweissstahl). 

3 . That all compounds of iron, with its ordinary ingredients 
which have been cast from a fluid state into malleable masses, 
and which will not sensibly harden by being quenched in water 
while at a red heat, shall be called ingot-iron (Flusseisen). 

4 . That all such compounds, when they will, from any cause, 
so harden, shall be called ingot-steel (Flussstahl) . 

On the other hand, in English-speaking countries, those 
compounds which have been aggregated from a pasty mass, 
usually in the puddling-furnace, and which contain slag, are 
generally called wrought-iron, while those which have been cast 
from a molten state into a malleable mass are generally called steel. 

While this classification is not perfect, it states the most 
common practice in a general way. Exceptions, two of which 
are that it does not include the cases of cementation steel and 
of puddled steel, will not be discussed here. 

In view of the above, it will be plain that what is commonly 



44° APPLIED MECHANICS. 

called mild steel in America, would be called ingot-iron under 
the definition of the international committee. Steel is usually 
made by one of three processes, viz. : the crucible process, the 
Bessemer process, or the open-hearth process. 

While other processes, as the cementation process and others, 
are sometimes used, the three enumerated above are in most 
common use at the present time. 

Crucible Steel. — This is very commonly made by re-melting 
blister-steel in crucibles; the blister-steel being made by the 
cementation process, in which bars of very pure wrought-iron, 
especially low in phosphorus, are heated in contact with charcoal 
until they have absorbed the necessary amount of carbon. 

A cheaper process, and one much used at the present day, 
is to melt a mixture of charcoal and crude bar-iron in a 
crucible. 

Crucible steel, which is always high-carbon steel, is used for 
the finest cutlery, tools, etc., and wherever a very pure and 
homogeneous quality of steel is required. 

Bessemer Steel. — In the Bessemer process, a blast of air is 
blown into melted cast-iron, removing the greater part of its 
carbon and burning out more or less of the other ingredients. 
The process is conducted in a converter, which is usually so 
arranged that, when the operation is complete, it can be rotated 
around a horizontal axis to such an extent that the tuyeres are 
above the surface of the molten steel, and the blast is shut off. 

In the acid Bessemer process, the lining of the converter is 
made of some silicious substance, the burning of silicon being 
relied upon to develop a sufficiently high temperature to keep 
the metal fluid. 

In the basic Bessemer process, the lining of the converter is 
of such a nature as to resist the action of basic slags. It is usually 
made of dolomite, or of some kind of limestone. Burned lime is 
added to the charge to seize the silicon and phosphorus, the latter 
serving to develop a sufficiently high temperature. 



STEEL. 447 



In the latter part of the operation, the phosphorus is largely 
burned out, whereas in the acid process, in order to produce a 
steel that is low in phosphorus, it is necessary to use a pig-iron 
that is low in phosphorus. 

Open-hearth Steel. — In the open-hearth process, a charge of 
pig-iron and scrap is placed on the bed of a regenerative furnace, 
and exposed to the action of the flame, and is thus converted into 
steel. 

In the acid open-hearth process, the lining of the furnace is 
of a silicious nature, and is covered with sand, while in the basic 
it is usually of dolomite, or of some kind of limestone. 

Bessemer and open-hearth steel contain more impurities 
than crucible steel, but they are very much cheaper, and are 
just as suitable for many purposes. It is only in consequence 
of their introduction that steel can be extensively used on the 
large scale, as crucible steel would be too expensive for many 
purposes. 

Steel, unlike wrought-iron, is fusible; unlike cast-iron, it can 
be forged; and, with the exception of the harder grades, it can 
be welded by heating and hammering, the welding of high-carbon 
steel in large masses being a very uncertain operation, though 
small masses can be welded by taking proper care. 

The special characteristic, however, is, that, with the exception 
of the milder grades, when raised to a red heat and suddenly 
cooled, it becomes hard and brittle, and that, by subsequent 
heating and cooling, the hardness may be reduced to any desired 
degree. The first process is called hardening and the second 
tempering. 

The principal element in the steels that are ordinarily used is 
carbon; nevertheless, both Bessemer and open-hearth steel con- 
tain also silicon, manganese, sulphur, phosphorus, etc., which 
have more or less effect upon the resisting properties of the metal. 
Sulphur, silicon, and phosphorus usually come from the ore, 
the fuel, and the flux, while manganese, which is added, operates, 



44^ APPLIED MECHANICS. 

among other things, to render the steel ductile while hot, and 
therefore workable, and to absorb oxygen from the melted mass. 

Sulphur is injurious by causing brittleness when hot, and 
phosphorus by causing brittleness when cold. Phosphorus is 
the- most harmful ingredient in steel, so that when steel is to be 
used for structural purposes, it is important to have as little 
phosphorus as possible, and any excess of phosphorus is not to 
be tolerated. 

The injury done to steel plates by punching is greater than 
that done to iron plates: this injury can, however, be removed 
by annealing. Steel requires greater care in working it than 
iron, whether in punching, flanging, riveting, or other methods 
of working; otherwise it may, if overheated, burn, or receive 
other injury from careless workmanship. 

The chemical composition of steel is one important element 
in its resisting properties; but, on the other hand, the mode of 
working also has a great influence on the quality. 

The introduction of the Bessemer process was quickly fol- 
lowed by the general use of steel rails, and later, as this and the 
other processes for making steel for structural purposes have 
been developed, there has been a constant increase in the pur- 
poses for which steel has been used. 

One of the earlier applications was to the construction 
of steam-boilers, steel boiler-plate displacing almost entirely 
wrought-iron boiler-plate. Of late years the development of 
the steel manufacture has so perfected, and at the same time 
cheapened, structural steel that it is now used in most cases 
where wrought-iron was formerly employed. Thus the eye- bars 
and the struts of bridges are almost exclusively made of steel, 
also such shapes as angle-irons, channel-bars, Z bars, tee iron, 
I beams, etc., are almost exclusively made of steel, and while 
steel has long been used for many parts of machinery, never- 
theless it is now generally used in many cases where a con- 
siderable fear of it formerly existed, as in main rods, parallel 



STEEL. 449 



rods, and crank-pins, and in a large number of parts of machinery 
subjected to more or less vibration. On the other hand, the 
steel used for tools is, of course, high-carbon steel. 

Tools are almost always made of crucible steel, and they 
have of course a high percentage of carbon, a high tensile strength, 
and especially should they be capable of being well hardened 
and taking a good temper. 

The usual steel of commerce may be called carbon steel, 
because, although it always contains small percentages of other 
ingredients, nevertheless carbon is the ingredient that princi- 
pally determines its properties. When iron or steel is alloyed 
with large percentages of certain substances, the resulting al- 
loys enjoy certain special properties, and these alloys still bear 
the name of steel. Two of the most prominent of these are 
manganese steel and nickel steel. 

Regarding the first it may be said that although carbon steel 
becomes practically useless when the manganese reaches about 
i J per cent, nevertheless with manganese exceeding about 7 per 
cent we obtain manganese steel which is so hard that it is exceed- 
ingly difficult to machine it. 

The alloy that has come into most prominent notice recently 
is nickel steel, which cons&ts most commonly of a carbon steel 
with from 0.2 to 0.4 per cent of carbon and with from 3 to 5 
per cent of nickel. With this amount of nickel the tensile strength 
is very much increased, but more especially is the limit of elasticity 
increased by a very large amount; and while the contraction of 
area at fracture and the ultimate elongation per cent are a little 
less than that of carbon steel with the same percentage of carbon, 
they are not less than those of carbon steel of the same tensile 
strength. 

It is used for armor-plates, for which it is specially suitable 
on account of the fact that the nickel renders the steel more 
sensitive to hardening. It is finding, also, a great many other 
uses to which it is specially adapted by its peculiar properties. 



450 APPLIED MECHANICS. 

It has been used for bicycle-spokes, for shafts for ocean steam- 
ships, for piston-rods, and for various other purposes. Among 
the many examples given by Mr. D. H. Browne in a paper before 
the American Institute of Mining Engineers is a case where the 
presence of 3.5 per cent nickel increased the ultimate strength 
of 0.2 per cent carbon steel from 55000 to 85000, and the elastic 
limit from 28000 to 48000 pounds per square inch, while the 
contraction of area at fracture was only decreased from 60 per 
cent to 55 per cent. 

The quality of steel to be used for different purposes differs, 
and while the specifications for any one purpose, made by different 
engineers, and by different engineering societies, often differ, the 
work of the American Society for Testing Materials is tending 
to harmonize them as far as possible. The result of their efforts 
is shown in the following set of specifications. 

AMERICAN SOCIETY FOR TESTING MATERIALS. 
SPECIFICATIONS FOR STEEL. 

STEEL CASTINGS. 

Adopted 1 90 1. Modified 1905. 

Process of Manufacture. 

1. Steel for castings may be made 1§y the open-hearth, crucible, 
or Bessemer process. Castings to be annealed unless otherwise 
specified. 

Chemical Properties. 

2. Ordinary castings, those in which no physical requirements are 
Ordinary specified, shall not contain over 0.40 per cent of carbon, 

Castings. nor oyer aQ g p er cent Q £ phosphorus. 

3. Castings which are subjected to physical test shall not contain 
Tested over °-°5 P er cent °f phosphorus, nor over 0.05 per cent 

Castings. of sulphur< 

Physical Properties. 

4. Tested castings shall be of three classes: "hard," "medium, " 
Tensile an d "soft." The minimum physical qualities required in 
Tests# each class shall be as follows : 



STEEL CASTINGS. 



451 



Tensile strength, pounds per square inch . 
Yield-point, pounds per square inch . 
Elongation, per cent in 2 inches 
Contraction of area, per cent 



Hard 
Castings. 


Medium 
Castings. 


Soft 

Castings. 


85000 
38250 

15 
20 


70000 

315°° 
18 

25 


60000 

27000 

22 

30 



5. A test to destruction may be substituted for the tensile test in 
the case of small or unimportant castings by selecting 

three castings from a lot. This test shall show the material 
to be ductile and free from injurious defects and suitable for the pur- 
poses intended. A lot shall consist of all castings from the same melt 
or blow, annealed in the same furnace charge. 

6. Large . castings are to be suspended and hammered all over. 
No cracks, flaws, defects, nor weakness shall appear after Percussive 
such treatment. Test * 

7. A specimen one inch by one-half inch (i"Xi") shall bend cold 
around a diameter of one inch (1") without fracture on Bending 
outside of bent portion, through an angle of 120 for "soft" Test ' 
castings and of 90 for "medium " castings. 

Test Pieces and Methods of Testing. 

8. The standard turned test specimen one-half inch (J") diameter 
and two inch (2") gauged length shall be used to determine Test Speci- 
the physical properties specified in paragraph No. 4. It Tensile Test.. 
is shown in Fig. 1. (See page 398.) 

9. The number of standard test specimens shall depend upon the 
character and importance of the castings. A test piece 

shall be cut cold from a coupon to be moulded and cast LoStk^of 
on some portion of one or more castings from each melt JjjJInJf Spec ~ 
or blow or from the sink-heads (in case heads of sufficient 
size are used). The coupon or sink-head must receive the same treat- 
ment as the casting or castings before the specimen is cut out, and 
before the coupon or sink-head is removed from the casting. 

10. One specimen for bending test one inch by one-half inch (i"X J") 
shall be cut cold from the coupon or sink-head of the cast- Test Specimen 
ing or dastings as specified in paragraph No. 9. The for Bendin s- 
bending test may be made by pressure or by blows. 



452 



APPLIED MECHANICS, 



YieId=point. 



ii. The yield -point specified in paragraph No. 4 shall be deter- 
mined by the careful observation of the drop of the beam 
or halt in the gauge of the testing-machine. 

12. Turnings from tensile specimen, drillings from the bending 
specimen, or drillings from the small test ingot, if preferred 

Sample for , r , . ' , ,, , 1 . 1 . • n , 

Chemical by the inspector, shall be used to determine whether or 

Ansi vsis 

not the steel is within the limits in phosphorus and sulphur 



specified in paragraphs Nos. 2 and 3. 

Finish. 

13. Castings shall be true to pattern, free from blemishes, flaws, 
or shrinkage cracks. Bearing-surfaces shall be solid, and no porosity 
shall be allowed in positions where the resistance and value of the 
casting for the purpose intended will be seriously affected thereby. 

Inspection. 

14. The inspector, representing the purchaser, shall have all 
reasonable facilities afforded to him by the manufacturer to satisfy 
him that the finished material is furnished in accordance with these 
specifications. All tests and inspections shall be made at the place of 
manufacture, prior to shipment. 

STEEL FORGINGS. 

Adopted 190 1. Modified 1905. 

Process of Manufacture. 

1. Steel for forgings may be made by the open-hearth, crucible, or 
Bessemer process. 

Chemical Properties. 

2. There shall be four classes of steel forgings which shall conform 
to the following limits in chemical composition: 





Forgings 


Forgings 


Forgings of 


Loco- 


Forgings of 
Nickel Steel, 




of Soft or 


of Carbon 


Carbon Steel 


motive 




Low-car- 


Steel not 


Oil-tempered 


Forg- 


Oil-tempered 




bon Steel. 


Annealed. 


or Annealed. 


ings. 


or Annealed. 




Per Cent. 


Per Cent. 


Per Cent. 


PerCt, 


Per Cent. 


Phosphorus shall not exceed 


. 10 


0.06 


0.04 


0.05 


0.04 


Sulphur 


0. 10 


0.06 


0.04 


0.05 


0.04 


Manganese " " " 


— 


— 


— 


0.60 


— 


Nickel 


— 


— 


— 


3 . to 4 . 



Tensile 
Tests. 



Physical Properties. 
3. The minimum physical qualities required of the 
different-sized forgings of each class shall be as follows: 



STEEL FORGINGS. 



453 



Tensile 


Yield- 


Elonga- 


Contrac- 




Strength. 


point . 


tion in 2 


tion of 




Lbs. per 


Lbs. per 


Inches. 


Area. 




Sq. In. 


Sq. In. 


Per Cent. 


Per Cent. 












Soft Steel or Low-carbon Steel. 


58000 


29000 


28 


35 


For solid or hollow f orgings, no diameter 
or thickness of section to exceed 10". 

Carbon Steel not Annealed. 


75000 


375°° 

Elastic 
Limit. 


18 


30 


For solid or hollow f orgings, no diameter 
or thickness of section to exceed 10". 

Carbon Steel Annealed. 


80000 


40000 


22 


35 


For solid or hollow f orgings, no diameter 
or thickness of section to exceed 10". 


75000 


375°° 


23 


35 


For solid f orgings, no diameter to exceed 
20" or thickness of section 15". 


70000 


35000 


24 


30 


For solid forgings, over 20" diameter. 
Carbon Steel Oil-tempered. 


90000 


55°°° 


20 


45 


For solid or hollow forgings, no diameter 
or thickness of section to exceed 3". 


85000 


50000 


22 


45 


For solid forgings of rectangular sections 
not exceeding 6" in thickness or hol- 
low forgings, the walls of which do not 
exceed 6" in thickness. 


80000 


45000 


23 


40 


For solid forgings of rectangular sections 
not exceeding 10" in thickness or hol- 
low forgings, the walls of which do not 
exceed 10" in thickness. 


80000 


40000 


20 


25 


Locomotive Forgings. 
Nickel Steel Annealed. 


80000 


50000 


25 


45 


For solid or hollow forgings, no diameter 
or thickness of section to exceed 10". 


80000 


45000 


25 


45 


For solid forgings, no diameter to exceed 
20 " or thickness of section 15". 


80000 


45000 


24 


40 


For solid forgings, over 20" diameter. 
Nickel Steel, Oil-tempered. 


95000 


65000 


21 


50 


For solid or hollow forgings, no diameter 
or thickness of section to exceed 3". 


90000 


60000 


22 


50 


For solid forgings of rectangular sections 
not exceeding 6" in thickness or hol- 
low forgings, the walls of which do not 
exceed 6" in thickness. 


85000 


55000 


24 


45 


For solid forgings of rectangular sections 
not exceeding 10" in thickness or hol- 
low forgings, the walls of which do not 
exceed 10" in thickness. 



454 APPLIED MECHANICS. 

4. A specimen one inch by one-half inch (V'Xj") shall bend 

cold 180 without fracture on outside of the bent portion, 
Test - as follows: 

Around a diameter of \ ", for forgings of soft steel. 

Around a diameter of 1 J", for forgings of carbon steel not annealed. 

Around a diameter of ij", for forgings of carbon steel annealed, if 
20" in diameter or over. 

Around a diameter of 1", for forgings of carbon steel annealed^ 
if under 20" diameter. 

Around a diameter of 1", for forgings of carbon steel, oil tempered. 

Around a diameter of J", for forgings of nickel steel annealed. 

Around a diameter of 1", for forgings of nickel steel, oil tempered. 

For locomotive forgings no bending tests will be required. 

Test Pieces and Methods of Testing. 

5. The standard turned test specimen, one-half inch (J") diameter 
Test Sped- and two (2") gauged length, shall be used to determine 
siie Test. " the physical properties specified in paragraph No. 3. 

It is shown in Fig. 1. (See page 398.) 

6. The number and location of test specimens to be taken from 

a melt, blow, or a forging, shall depend upon its character 
Location a of and importance, and must therefore be regulated by 
imensf Spe<> individual cases. The test specimens shall be cut cold 

from the forging or full-sized prolongation of same parallel 
to the axis of the forging and half-way between the centre and outside, 
the specimens to be longitudinal; i.e., the length of the specimen to 
correspond with the direction in which the metal is most drawn out 
or worked. When forgings have large ends or collars, the test specimens 
shall be taken from a prolongation of the same diameter or section as 
that of the forging back of the large end or collar. In the case of 
hollow shafting, either forged or bored, the specimen shall be taken within 
the finished section prolonged, half-way between the inner and outer 
surface of the wall of the forging. 

7. The specimen for bending test one inch by one-half inch 
Tests cimen (^'Xj") shall be cut as specified in paragraph No. 6„ 
for Bending. The bending test may be made by pressure or by blows. 



OPEN-HEARTH BOILER PLATE AND RIVET STEEL. 455 

8. The yield -point specified in paragraph No. 3 shall be determined 
by the careful observation of the drop of the beam, or 

halt in the gauge of the testing machine. ie ~ poin 

9. The elastic limit specified in paragraph No. 3 shall be determined 
by means of an extensometer, which is to be attached to EIastic 

the test specimen in such manner as to show the change Limit. 

in rate of extension under uniform rate of loading, and will be taken 

at that point where the proportionality changes. 

10. Turnings from the tensile specimen or drillings from the bend- 
ing specimen or drillings from the small test ingot, if pre- 
ferred by the inspector, shall be used to determine whether chemical" 
or not the steel is within the limits in chemical composition na ysis * 
specified in paragraph No. 2. 

Finish. 

11. Forgings shall be free from cracks, flaws, seams, or other 
injurious imperfections, and shall conform to the dimensions shown 
on drawings furnished by the purchaser, and be made and finished in 
a workmanlike manner. 

Inspection. 

12. The inspector, representing the purchaser, shall have all reason- 
able facilities afforded him by the manufacturer to satisfy him that 
the finished material is furnished in accordance with these specifications. 
All tests and inspections shall be made at the place of manufacture, 
prior to shipment. 

OPEN-HEARTH BOILER PLATE AND RIVET STEEL. 

Adopted 1 90 1. 

Process of Manufacture. 

1. Steel shall be made by the open-hearth process. 

Chemical Properties. 

2. There shall be three classes of open-hearth boiler plate and 
rivet steel; namely, flange, or boiler steel, fire-box steel, and extra- 
soft steel, which shall conform to the following limits in chemical 
composition: 



456 



APPLIED MECHANICS. 





Flange or 

Boiler Steel. 

Per Cent. 


Fire-box 

Steel. 
Per Cent. 


Extra Soft 

Steel. 
Per Cent. 


Phosphorus shall not exceed . 

Sulphur " " " . . 
Manganese 


f Acid O.06 

\ Basic . 04 

0.05 

. 30 to . 60 


Acid . 04 

Basic 0.03 

0.04 

. 30 to . 50 


Acid . 04 

Basic 0.04 

0.04 

. 30 to . 50 







Boiler=rivet 
Steel. 



3. Steel for boiler rivets shall be of the extra-soft 
class as specified in paragraphs Nos. 2 and 4. 



Tensile 
Tests. 



Physical Properties. 
4. The three classes of open-hearth boiler plate and rivet steel — 
namely, flange or boiler steel, fire-box steel, and extra- 
soft steel — shall conform to the following physical qual- 



ities ; 





Flange or 
Boiler Steel. 


Fire-box 
Steel. 


Extra Soft 
Steel. 


Tensile strength, pounds 
per square inch . 

Yield-point, in pounds per 
square inch, shall not be 
less than 

Elongation, per cent in 8 
inches shall not be less 
than 


55000 to 65000 
£T. S. 

25 


52000 to 62000 

|T. S. 

26 


45000 to 55000 

*T. S. 

28 



5. For material less than five-sixteenths inch (&") and more than 

three-fourths inch (}") in thickness the following modifica- 
Modifications tions shall be made in the requirements for elongation : 

in Elongation ^ < ° 

£?. r .T hi . nand . . (a) For each increase of one-eighth inch (¥') in thick- 
Thick Material. v J y vo 7 

ness above three-fourths inch (f ) a deduction of one 

per cent (1%) shall be made from the specified elongation. 

(b). For each decrease of one-sixteenth inch (^ r/ ) in thickness 

below five-sixteenths inch (V) a deduction of two and one-half per 

cent (2^%) shall be made from the specified elongation. 

6. The three classes of open-hearth boiler plate and rivet steel 
Bendin shall conform to the following bending tests; and for this 
Tests. purpose the test specimen shall be one and one-half inches 
(ij") wide, if possible, and for all material three-fourths inch (}") or 
less in thickness the test specimen shall be of the same thickness as that 



OPEN-HEARTH BOILER PLATE AND RIVET STEEL, 45/ 

of the finished material from which it is cut, but for material more than 
three-fourths inch £") thick the bending-test specimen may be one- 
half inch (£") thick: 

Rivet rounds shall be tested of full size as rolled. 

(c). Test specimens cut from the rolled material, as specified above, 
shall be subjected to a cold bending test, and also to a quenched bending 
test. The cold bending test shall be made on the material in the con- 
dition in which it is to be used, and prior to the quenched bending test 
the specimen shall be heated to a light cherry-red, as seen in the dark, 
and quenched in water the temperature of which is between 8o° and 
90 Fahrenheit. 

(d). Flange or boiler steel, fire-box steel, and rivet steel, both before 
and after quenching, shall bend cold one hundred and eighty degrees 
(180 ) flat on itself without fracture on the outside of the bent portion. 

7. For fire-box steel a sample taken from a broken tensile-test 
specimen shall not show any single seam or cavity more Homogeneity 
than one-fourth inch (J") long in either of the three fractures Tests * 
obtained on the test for homogeneity as described below in paragraph 12. 

Test Pieces and Methods of Testing. 

8. The standard specimen of eight inch (8") gauged length shall be 
used to determine the physical properties specified in 
paragraphs Nos. 4 and 5. The standard shape of the men for 
test specimen for sheared plates shall be as shown in Fig. 

2. (See page 398.) 

For other material the test specimen may be the same as for 
sheared plates, or it may be planed or turned parallel throughout its 
entire length ; and in all cases, where possible, two opposite sides of the 
test specimens shall be the rolled surfaces. Rivet rounds and small 
rolled bars shall be tested of full size as rolled. 

9. One tensile-test specimen will be furnished from each plate as 
it is rolled, and two tensile-test specimens will be furnished 

from each melt of rivet rounds. In case any one of these xenSe Tests. 
develops flaws or breaks outside of the middle third of its 
gauged length, it may be discarded and another test specimen sub- 
stituted therefor. 



45^ APPLIED MECHANICS. 

10. For material three-fourths inch (}") or less in thickness the 

bending-test specimen shall have the natural rolled surface 

mens for on two opposite sides. The bending-test specimens cut 

Bending. f rom pi ates g^jj foe Qne an( j one _ na lf i ncnes (ij") wide, 

and for material more than three-fourths inch (}") thick the bending- 
test specimens may be one-half inch (i") thick. The sheared edges 
of bending-test specimens may be milled or planed. The bending- 
test specimens for rivet rounds shall be of full size as rolled. The 
bending test may be made by pressure 'or by blows. 

ii. One cold-bending specimen and one quenched -bending specimen 
will be furnished from each plate as it is rolled. Two 
Bending cold-bending specimens and two quenched -bending speci- 
mens will be furnished from each melt of rivet rounds. 
The homogeneity test for fire-box steel shall be made on one of the 
broken tensile-test specimens. 

12. The homogeneity test for fire-box steel is made as follows: A 

portion of the broken tensile-test specimen is either nicked 
Homogeneity with a chisel or grooved on a machine, transversely about 
Fire=box a sixteenth of an inch {h") deep, in three places about 

two inches (2") apart. The first groove should be made 
on one side, two inches (2") from the square end of the specimen; 
the second, two inches (2") from it on the opposite side; and the third, 
two inches (2") from the last, and on the opposite side from it. The 
test specimen is then put in a vise, with the first groove about a quarter 
of an inch (\") above the jaws, care being taken to hold it firmly. The 
projecting end of the test specimen is then broken off by means of a 
hammer, a number of light blows being used, and the bending being 
away from the groove. The specimen is broken at the other two 
grooves in the same way. The object of this treatment is to open 
and render visible to the eye any seams due to failure to weld up, or to 
foreign interposed matter, or cavities due to gas bubbles in the ingot. 
After rupture, one side of each fracture is examined, a pocket lens 
being used, if necessary, and the length of the seams and cavities is 
determined. 

13. For the purposes of this specification the yield-point shall be 

determined by the careful observation of the drop of the 
ie -point. k eam or k a it in the gauge of the testing machine. 



OPEN-HEARTH BOILER PLATE AND RIVET STEEL. 459 

14. In order to determine if the material conforms to the chemical 
limitations prescribed in paragraph 2 herein, analysis 

shall be made of drillings taken from a small test ingot, chemical 01 " 
An additional check analysis may be made from a tensile Ana,ys,s * 
specimen of each melt used on an order, other than in locomotive 
fire-box steel. In the case of locomotive fire-box steel a check analysis 
may be made from the tensile specimen from each plate as rolled. 

Variation in Weight. 

15. The variation in cross section or weight of more than 2 \ per 
•cent from that specified will be sufficient cause for rejection, except in 
the case of sheared plates, which will be covered by the following per- 
missible variations: 

(e) Plates 12^ pounds per square foot for heavier, up to 100 inches 
wide when ordered to weight, shall not average more than 2% per cent 
variation above or 2 J per cent below the theoretical weight. When 
100 inches wide and over, 5 per cent above or 5 per cent below the 
theoretical weight. 

(/) Plates under 12^ pounds per square foot, when ordered to 
weight, shall not average a greater variation than the following: 

Up to 75 inches wide, 2\ per cent above or 2 J per cent below the theo- 
retical weight. Seventy -five inches wide up to 100 inches wide, 5 per cent 
above or 3 per cent below the theoretical weight. When 100 inches 
wide and over, 10 per cent above or 3 per cent below the theoretical 
weight. 

(g) For all plates ordered to gauge there will be permitted an average 
excess of weight over that corresponding to the dimensions on the order 
equal in amount to that specified in the following table: 

Table of Allowances for Overweight for Rectangular Plates 
when ordered to gauge. 

Plates will be considered up to gauge if measuring not over y-J-j- 
inch less than the ordered gauge. 

The weight of one cubic inch of rolled steel is assumed to be 0.2833 
pound. 



460 



APPLIED MECHANICS. 



Plates 


\ Inch and 


over in Thickness. 


Thickness of 


Width of Plate. 








Plate. 
Inch. 


Up to 75 

Inches. 

Per Cent. 


75 to 100 

Inches. 

Per Cent. 


Over 100 

Inches. 

Per Cent. 




10 
8 


14 

12 


18 
16 


t 


7 
6 


IO 

8 


13 
10 


f 
Overf 


5 

Ah 
4 
3i 


7 

6 

5 


9 

8 

6i 



Plates under £ Inch in Thickness. 



Thickness of 
Plate. 
Inch. 


Width of Plate. 


Up to 50 

Inches. 

Per Cent. 


50 Inches 

and Above. 

Per Cent. 


| up to ^ 

Wl " T6 


10 
8£ 
7 


15 
10 



Finish. 

16. All finished material shall be free from injurious surface defects 
and laminations, and must have a workmanlike finish. 

Branding. 

17. Every finished piece of steel shall be stamped with the melt 
number, and each plate and the coupon or test specimen cut from it 
shall be stamped with a separate identifying mark or number. Rivet 
steel may be shipped in bundles securely wired together with the melt 
number on a metal tag attached. 

Inspection. 

18. The inspector, representing the purchaser, shall have all reason- 
able facilities afforded to him by the manufacturer to satisfy him that 
the finished material is furnished in accordance with these specifica- 
tions. All tests and inspections shall be made at the place of manu- 
facture, prior to shipment. 



STRUCTURAL STEEL FOR BUILDINGS. 



461 



STRUCTURAL STEEL FOR BUILDINGS. 

Adopted 1 90 1. 

Process of Manufacture. 

1. Steel may be made by either the open-hearth or Bessemer process. 

Chemical Properties. 

2. Each of the two classes of structural steel for buildings shall 
not contain more than o. 10 per cent of phosphorus. 

Physical Properties. 

3. There shall be two classes of structural steel for buildings, — 
namely, rivet steel and medium steel, — which shall con- Classes, 
form to the following physical qualities: 

4. Tensile Tests. 





Rivet Steel. 


Medium Steel. 


Tensile strength, pounds per square inch . 
Yield -point, in pounds per square inch, shall 

not be less than 

Elongation, per cent in 8 inches shall not be 

less than 


50000 to 60000 

|T. S. 

26 


60000 to 70000 

iT. S. 

22 





5. For material less than five-sixteenths inch (&") and more than 
three-fourths inch (f ") in thickness the following modifica- 
tions shall be made in the requirements for elongation: Modifications 

(a) For each increase of one-eighth inch ft") in thick- fc^ffSS 11 
ness above three-fourths inch ({") a deduction of one Thick Matenal * 
per cent (1%) shall be made from the specified elongation. 

(b) For each decrease of one-sixteenth inch (js") in thickness 
below five-sixteenths inch (&") a deduction of two and one-half per 
cent (2%%) shall be made from the specified elongation. 

(c) For pins the required elongation shall be five per cent (5%) 
less than that specified in paragraph No. 4, as determined on a test 
specimen the centre of which shall be one inch (1") from the surface. 

6. The two classes of structural steel for buildings shall conform 
to the following bending tests ; and for this purpose the test Bending 
specimen shall be one and one-half inches (ij") wide, if Tests - 
possible, and for all material three-fourths (} ") or less in thickness the test 
specimen shall be of the same thickness as that of the finished material 



462 APPLIED MECHANICS, 

from which it is cut, but for material more than three-fourths inch (f ") 
thick the bending-test specimen may be one-half inch (J") thick. 
Rivet rounds shall be tested of full size as rolled. 

(d) Rivet steel shall bend cold 180 flat on itself without fracture 
on the outside of the bent portion. 

(e) Medium steel shall bend cold 180 around a diameter equal 
to the thickness of the specimen tested, without fracture on the outside 
of the bent portion. 

Test Pieces and Methods of Testing. 

7. The standard test specimen of eight-inch (8") gauged length 

shall be used to determine the physical properties specified 
men for Ten- in paragraphs Nos. 4 and 5. The standard shape of the 

test specimen for sheared plates shall be as shown by 
Fig. 2. (See page 398.) For other material the test specimen may be 
the same as for sheared plates or it may be planed or turned parallel 
throughout its entire length and, in all cases where possible, two oppo- 
site sides of the test specimen shall be the rolled surfaces. Rivet 
rounds and small rolled bars shall be tested of full size as rolled. 

8. One tensile-test specimen shall be taken from the finished 
Number of material of each melt or blow; but in case this develops 
Tensile Tests. fl awSj or breaks outside of the middle third of its gauged 
length, it may be discarded and another test specimen substituted 
therefor. 

9. One test specimen for bending shall be taken from the finished 

material of each melt or blow as it comes from the rolls, and 
men for for material three-fourths inch (}") and less in thickness 

this specimen shall have the natural rolled surface on two 
opposite sides. The bending-test specimen shall be one and one- 
half inches (ii") wide, if possible; and for material more than three- 
fourths inch (f") thick the bending-test specimen may be one-half 
inch (J") thick. The sheared edges of bending-test specimens may 
be milled or planed. 

Rivet rounds shall be tested of full size as rolled. 

(/) The bending test may be made by pressure or by blows. 

10. Material which is to be used without annealing or further 
Annealed treatment shall be tested for tensile strength in the con- 
mens. dition in which it comes from the rolls. Where it is 



STRUCTURAL STEEL FOR BULLDINGS. 463 

impracticable to secure a test specimen from material which has 
been annealed or otherwise treated, a full-sized section of tensile- 
test specimen length shall be similarly treated before cutting the tensile- 
test specimen therefrom. 

11. For the purposes of this specification the yield-point shall be 
determined by the careful observaton of the drop of the Yield-point. 
beam or halt in the gauge of the testing machine. 

12. In order to determine if the material conforms 

to the chemical limitations prescribed in paragraph No. 2 chemical 01 * 
herein, analysis shall be made of drillings taken from a na,ys,s * 
small test ingot. 

Variation in Weight. 

13. The variation in cross section or weight of more than 2 J per 
cent from that specified will be sufficient cause for rejection, except in 
the case of sheared plates, which will be covered by the following per- 
missible variations : 

(g) Plates 12 J pounds per square foot or heavier, up to 100 inches 
wide, when ordered to weight, shall not average more than 2 J per 
cent variation above or 2J per cent below the theoretical weight. 
When 100 inches wide, and over 5 per cent above or 5 per cent below 
the theoretical weight. 

(h) Plates under 12^ pounds per square foot, when ordered to 
weight, shall not average a greater variation than the following: 

Up to 75 inches wide, 2 J per cent above or 2\ per cent below the 
theoretical weight. Seventy-five inches wide up to 100 inches wide, 5; 
per cent above or 3 per cent below the theoretical weight. When 100 
inches wide and over, 10 per cent above or 3 per cent below the 
theoretical weight. 

(i) For all plates ordered to gauge, there will be permitted an 
average excess of weight over that corresponding to the dimensions 
on the order equal in amount to that specified in the following table : 
Table of Allowances for Overweight for Rectangular Plates 
when ordered to gauge. 

Plates will be considered up to gauge if measuring not over j-J^ 
inch less than the ordered gauge. 

The weight of 1 cubic inch of rolled steel is assumed to be 0.2833 
pound. 



464 



APPLIED MECHANICS. 



Plates \ Inch and over in Thickness. 



Thickness of 
Plate. 
Inch. 


Width of Plate. 


Up to 75 

Inches. 

Per Cent. 


75 to 100 

Inches. 
Per Cent. 


Over 100 

Inches. 

Per Cent. 


t 

I 

1 
Over f 


IO 
8 

7 
6 

li 

4 
3* 


14 
12 
10 

8 

7 

6* 

6 

5 


18 
16 

13 
IO 

9 

8* 
8 
6§ 



Plates under \ Inch in Thickness. 



Thickness of 
Plate. 
Inch. 


Width of Plate. 


Up to 50 

Inches. 

Per Cent. 


50 Inches 

and Above. 

Per Cent. 


£• up to ^ 


IO 
8* 
7 .- 


15 

12* 

10 



Finish. 

14. Finished material must be free from injurious seams, flaws, or 
cracks, and have a workmanlike finish. 

Branding. 

15. Every finished piece of steel shall be stamped with the melt or 
blow number, except that small pieces may be shipped in bundles 
securely wired together with the melt or blow number on a metal tag 
attached. . 

Inspection. 

16. The inspector, representing the purchaser, shall have all reason- 
able facilities accorded to him by the manufacturer to satisfy him that 
the finished material is furnished in accordance with these specifications. 
All tests and inspections shall be made at the place of manufacture,., 
prior to shipment. 






SPECIFICATIONS FOR STEEL FOR BRIDGES. 



465 



STRUCTURAL STEEL FOR BRIDGES. 

Adopted 1905. 

1. Steel shall be made by the open-hearth process. Manufacture. 

2. The chemical and physical properties shall conform chemical and 

to the following limits: Properties. 



Elements Considered. 

Phosphorus Max. { ^ c 

Sulphur Max 

Ult. tensile strength 

Pounds per sq. in 

Elong.: Min. per cent, in 8 in, 
(Fig. 1) 

Elong.: Min. per cent, in 2 in, 

(Fig. 1) 

Character of fracture 

Cold bend without fracture 



Structural Steel. 



0.04 per cent. 

0.08 

0.05 



Desired 
60,000 

f 1,500,000* 
\ Ult. tens, str, 

22 
Silky 

180 flat f 



Rivet Steel. 



0.04 per cent. 

0.04 

0.04 



Desired 
65,000 

1,500,000 
Ult. tens. str. 



Silky 
180 flat % 



Steel Castings. 



0.05 per 

0.08 

0.05 


cent. 
< « 



Not less than 
65,000 



18 
Silky or fine 

granular. 
90 . d= 3 t 



Retests. 



* See par. 11. f See par. 12, 13 and 14. { See par. 15. 

The yield-point, as indicated by the drop of beam, shall be recorded 
in the test reports. 

3. If the ultimate strength varies more than 4,000 lbs. from that 
desired, a retest may be made, at the discretion of the inspec- 
tor, on the same gauge, which, to be acceptable, shall be 
within 5,000 lbs. of the desired ultimate. 

4. Chemical determinations of the percentages of carbon, phos- 
phorus, sulphur, and manganese shall be made by the Chemical r^, 
manufacturer from a test ingot taken at the time of the terminations. 
pouring of each melt of steel and a correct copy of such analysis shall 
be furnished to the engineer or his inspector. Check analyses shall be 
made from finished material, if called for by the purchaser, in which 
case an excess of 25 per cent above the required limits will be allowed. 

5. Speciemns for tensile and bending tests for plates, shapes, and 
bars shall be made by cutting coupons from the finished piatej . Sh 
product, which shall have both faces rolled and both edges and Bars. 
milled to the form shown by Fig. 2, page 398; or with both edges 



466 APPLIED MECHANICS. 

parallel; or they may be turned to a diameter of } inch for a length 

of at least 9 inches, with enlarged ends. 

Rivets. 6. Rivet rods shall be tested as rolled. 

7. Specimens shall be cut from the finished rolled or forged bar in 
Pins and suc ^ manner tnat tne centre of the specimen shall be 1 
Rollers. mcn from the surface of the bar. The specimen for tensile 
test shall be turned to the form shown by Fig. 1, page 398. The 
specimen for bending test shall be 1 inch by \ inch in section. 

8. The number of tests will depe*id on the character and import- 
steel Cast- ance °^ ^ e castm g s - Specimens shall be cut cold from 
ings. coupons moulded and cast on some portion of one or more 
castings from each melt or from the sink-heads, if the heads are of 
sufficient size. The coupon or sink-head, so used, shall be annealed 
with the casting before it is cut off. Test specimens to be of the form 
prescribed for pins and rollers. 

9. Material which is to be used without annealing or further treat- 
Conditions m ent shall be tested in the condition in which it comes 
for Tests. from the rolls. When material is to be annealed or other- 
wise treated before use, the specimens for tensile tests, representing 
such material, shall be cut from properly annealed or similarly treated 
short lengths of the full section of the bar. 

10. At least one tensile and one bending test shall be made from 

Number of eac ^ me ^ °^ stee ^ as ro ^ e d- I* 1 case stee * differing f inch 
Tests. an d more in thickness is rolled from one melt, a test shall be 

made from the thickest and thinest material rolled. 

11. For material less than 5-16 inch and more than } inch in 

thickness the following modifications will be allowed in the 
onga on. requirements for elongation: 

(a) For each 1-16 inch in thickness below 5-16 inch, a deduction 

of 2 J will be allowed from the specified percentage. 

(b) For each J inch in thickness above f inch, a deduction of 1 

will be allowed from the specified percentage. 

12. Bending tests may be made by pressure or by blows. Plates, 
Bending shapes, and bars less than 1 inch thick shall bend as called 
Tests. f or m paragraph 2. 

13. Full-sized material for eye-bars and other steel 1 inch thick 



SPECIFICATIONS FOR STEEL FOE BRIDGES. 467 

and over, tested as rolled, shall bend cold 180 around _ „ . . 

' ' m Fuli=sized 

a pin the diameter of which is equal to twice the thickness Bends 
of the bar, without a fracture on the outside of bend. 

14. Angles f inch and less in thickness shall open flat, and angles 
\ inch and less in thickness shall bend shut, cold, under -, . 

6 - ' ' I ests on 

blows of a hammer, without sign of fracture. This test Angles, 
will be made only when required by the inspector. 

15. Rivet steel, when nicked and bent around a bar of the same 
diameter as the rivet rod, shall give a gradual break and a Tests on 
fine, silky, uniform fracture. Rivet steel - 

16. Finished material shall be free from injurious seams, flaws, 
cracks, defective edges, or other defects, and have a smooth 
uniform, workmanlike finish. Plates 36 inches in width Fm,sh * 
and under shall have rolled edges. 

17. Every finished piece of steel shall have the melt number and 
the name of the manufacturer stamped or rolled upon it. M .. 
Steel for pins and rollers shall be stamped on the end. 

Rivet and lattice steel and other small parts may be bundled with the 
above marks on an attached metal tag. 

18. Material which, subsequent to the above tests at the mills and 
its acceptance there, develops weak spots, brittleness, 

cracks or other imperfections, or is found to have injurious ejec lons * 
defects, will be rejected at the shop and shall be replaced by the manu- 
facturer at his own cost. 

19. A variation in cross-section or weight of each piece of steel 
of more than 2J per cent from that specified will be suffi- p ermiss j We 
cient cause for rejection, except in case of sheared plates, Variations, 
which will be covered by the following permissible variations, which 
are to apply to single plates. 

When Ordered to Weight. 

20. Plates \2\ pounds per square foot or heavier: Variations! 

(a) Up to 100 inches wide, 2 J per cent above or below the pre- 

scribed weight. 

(b) One hundred inches wide and over, 5 per cent above or below. 

21. Plates under i2§ pounds per square foot: 

(a) Up to 75 inches wide, 2 J per cent above or below. 



468 



APPLIED MECHANICS. 



(b) Seventy-five inches and up to ioo inches wide, 5 per cent above 
or 3 per cent below. 

(c) One hundred inches wide and over, 10 per cent above ©r 3 
per cent below. 

When Ordered to Gauge. 

22. Plates will be accepted if they measure not more 
than 0.01 inch below the ordered thickness. 

23. An excess over the nominal weight corresponding to the dimen- 
sions on the order, will be allowed for each plate, if not more than that 
shown in the following tables, one cubic inch of rolled steel being 
assumed to weigh 0.2833 pound. 

24. Plates J inch and over in thickness. 



Permissible 
Variations. 









Width of Plate. 




Thickness 
Ordered. 


Nominal 
Weights. 




















Up to 75". 


75" and up 
to 100". 


100" and up 
to 115'. 


Over us". 


1-4 inch. 
5-16 » 


10.20 lbs. 
12-75 " 


10 per cent. 
8 " " 


14 per cent. 
12 " " 


18 per cent. 
16 " " 




3-8 " 
7-16 " 


I5-30 " 
I7-85 " 


7 " " 
6 " " 


10 " " 

8 " " 


13 " " 
10 " " 


17 per cent. 
13 " " 


1-2 " 
9-16 " 

5-8 " 
Over 5-8 " 


20 . 40 " 
22.95 " 
25-50 " 


5 " " 
4i " " 
4 « « 


7 " " 

6 " " 

5 " " 


9 «« « 

8 " " 
6J '"' " 


12 *' " 
11 M " 
10 " " 

9 » » 





25. Plates under J inch in thickness. 



Thickness Ordered. 


Nominal Weights. 
Pounds per 
Square Feet. 


Width of Plate. 


Up to 50". 


50" and up to 
70". 


Over 70". 


1-8" up to 5-32" 

5-32" " 3-16" 

3-1 6'" '< 1-4" 


5.10 to 6.37 

6-37 " 7-65 
7.65 " 10.20 


10 per cent. 

a* ;; ;; 

7 


15 percent. 

mi "* tl 

10 " " 


20 per cent. 
17 " » 
15 " " 



26. 



The purchaser shall be furnished complete copies of mill orders 
and no material shall be rolled, nor work done, before the 
purchaser has been notified where the orders have been 
placed, so that he may arrange for the inspection. 



Inspection 
and Testing. 



STRUCTURAL STEEL FOR SHIPS. 



469 



27. The manufacturer shall furnish all facilities for inspecting and 
testing the weight and quality of all material at the mill where it is 
manufactured. He shall furnish a suitable testing machine for testing 
the specimens, as well as prepare the pieces for the machine, free of cost. 

28. When an inspector is furnished by the purchaser to inspect 
material at the mills, he shall have full access, at all times, to all parts 
■of mills where material to be inspected by him is being manufactured. 



STRUCTURAL STEEL FOR SHIPS. 

Adopted 1 90 1 for bridges and ships. Restricted to ships, 1905. 

Process of Manufacture. 

1. Steel shall be made by the open-hearth process. 

Chemical Properties. 

2. Each of the three classes of structural steel for ships shall con- 
form to the following limits in chemical composition: 





Steel Made by 
the Acid 
Process. 
Per Cent. 


Steel Made by 
the Basic 
Process. 
Per Cent. 


Phosphorus shall not exceed . 
Sulphur " " 


O.08 
O.06 


O.06 
0.06 



Physical Properties. 

3. There shall be three classes of structural steel for ships, — • 
namely, rivet steel, soft steel, and medium steel, — which 
shall conform to the following physical qualities: 

4. Tensile Tests. 



Classes. 





Rivet Steel. 


Soft Steel. 


Medium Steel. 


Tensile strength, pounds 
per square inch . 

Yield -point, in pounds per 
square inch shall not be 
less than 

Elongation, per cent in 8 
inches shall not be less 
than 


50000 to 60000 

§T.S. 

26 


52000 to 62000 

JT. S. 
25 


60000 to 70000 

JT.S. 
22 



470 APPLIED MECHANICS. 



5. For material less than five-sixteenths inch (&") and more than 

three-fourths inch (f") in thickness the following modifi- 
Modjfications cations shall be made in the requirements for elongation : 
for Thin and (a) For each increase of one-eighth inch (-£") in thickness 

Thick Material. , V ' , p , ." , „„ N f , . V8 , ' 

above three-fourths men (f ) a deduction of one per cent 
(1%) shall be made from the specified elongation. 

(b) For each decrease of one-sixteenth inch (ife") in thickness 
below five-sixteenths inch (&") a deduction of two and one-half per 
cent (2 \%) shall be made from the specified elongation. 

(c) For pins made from any of the three classes of steel the required 
elongation shall be five per cent (5%) less than that specified in para- 
gaph No. 4, as determined on a test specimen, the center of which shall 
be one inch (1") from the surface. 

6. Eye-bars shall be of medium steel. Full-sized tests shall show 
Tensile Tests l ^i per cent elongation in fifteen feet of the body of the 
ofEye=bars. eye-bar, and the tensile strength shall not be less than 
55,000 pounds per square inch. Eye-bars shall be required to break 
in the body; but, should an eye-bar break in the head, and show twelve 
and onerhalf per cent (12^%) elongation in fifteen feet and the tensile 
strength specified, it shall not be cause for rejection, provided that not 
more than one-third (J) of the total number of eye-bars tested break 
in the head. 

7. The three classes of structural steel for ships shall conform 
Bending to ^ e f° nowm g bending tests; and for this purpose 
Tests. the test specimen shall be one and one-half inches wide, 
if possible, and for all material three-fourths inch ($") or less in thick- 
ness the test specimen shall be of the same thickness as that of the 
finished material from which it is cut, but for material more than 
three-fourths inch (f") thick the bending-test specimen may be one- 
half inch (J") thick. 

Rivet rounds shall be tested of full size as rolled. 

(d) Rivet steel shall bend cold 180 flat on itself without fracture 
on the outside of the bent portion. 

(e) Soft steel shall bend cold 180 flat on itself without fracture on 
the outside of the bent portion. 

(/) Medium steel shall bend cold 180 around a diameter equal to 
the thickness of the specimen tested, without fracture on the outside of 
the bent portion. 



STRUCTURAL STEEL FOR SHIPS. 47 1 

Test Pieces and Methods of Testing. 

8. The standard test specimen of eight inch (8") gauged length shall 
be used to determine the physical properties specified in 

Test SdccI= 

paragraphs Nos. 4 and 5. The standard shape of the test men for Ten- 
specimen for sheared plates shall be as shown by Fig. 2, 
page 398. For other material the test specimen may be the same as 
for sheared plates, or it may be planed or turned parallel throughout 
its entire length; and, in all cases where possible, two opposite sides 
of the test specimens shall be the rolled surfaces. Rivet rounds and 
small rolled bars shall be tested of full size as rolled. 

9. One tensile-test specimen shall be taken from the finished material 
of each melt; but in case this develops flaws, or breaks Numberof 
outside of the middle third of its gauged length, it may Tensile Tests. 
be discarded, and another test specimen substituted therefor. 

10. One test specimen for bending shall be taken from the finished 
material of each melt as it comes from the rolls, and for 
material three-fourths inch (}") and less in thickness mensibr'" 
this specimen shall have the natural rolled surface on en Ing * 
two opposite sides. The bending-test specimen shall be one and one 
half inches (ij") wide, if possible, and for material more than three- 
fourths inch (f") thick the bending-test specimen may be one-half 
inch (J") thick. The sheared edges of bending-test specimens may 
be milled or planed. 

(g) The bending test may be made by pressure or by blows. 

11. Material which is to be used without annealing or further 
treatment shall be tested for tensile strength in the con- 
dition in which it comes from the rolls. Where it is imprac- Test Speci- 
ticable to secure a test specimen from material which m n 

has been annealed or otherwise treated, a full-sized section of tensile 
test, specimen length, shall be similarly treated before cutting the 
tensile-test specimen therefrom. 

12. For the purpose of this specification the yield-point shall be 
determined by the careful observation of the drop of the 

beam or halt in the gauge of the testing machine. 

13. In order to determine if the material conforms to 

the chemical limitations prescribed in paragraph No. 2 Chemical" 
herein, analysis shall be made of drillings taken from a n ysis * 
small test ingot. 



472 



APPLIED MECHANICS. 



Variation in Weight. 

14. The variation in cross section or weight of more than 2 J per 
cent from that specified will be sufficient cause for rejection, except in 
the case of sheared plates, which will be covered by the following per- 
missible variations: 

(h) Plates 12 J pounds per square foot or heavier, up to 100 inches 
wide, when ordered to weight, shall not average more than 2 J per cent 
variation above or 2J per cent below the theoretical weight. When 
100 inches wide and over, 5 per cent above or 5 per cent below the 
theoretical weight. 

(t) Plates under 12 J pounds per square foot, when ordered to weight, 
shall not average a greater variation than the following: 

Up to 75 inches wide, 2 J per cent above or 2j per cent below the 
theoretical weight. 75 inches wide up to 100 inches wide, 5 per cent 
above or 3 per cent below the theoretical weight. When 100 inches wide 
and over, 10 per cent above or 3 per cent below the theoretical weight. 

(/) For all plates ordered to gauge there will be permitted an average 
excess of weight over that corresponding to the dimensions on the order 
equal in amount to that specified in the following table: 

Table of Allowances for Overweight for Rectangular Plates 
when ordered to gauge. 

Plates will be considered up to gauge if measuring not over -fl- 
inch less than the ordered gauge. 

The weight of 1 cubic inch of rolled steel is assumed to be 0.2833 

pound. 

Plate I Inch and over in Thickness. 







Width of Plate 




Thickness of 














Plate. 








Inch. 


Up to 75 


75 to IOO 


Over 100 




Inches. 


Inches. 


Inches. 




Per Cent. 


Per Cent. 


Per Cent. 


\ 


IO 


14 


18 


& 


8 


12 


16 


f 


7 


IO 


13 


& 


6 


8 


IO 


1 


5 


7 


9 


A 


4* 


6* 


8* 


f 


4 


6 


8 


Overf 


3* 


5 


6| 



STEEL AXLES. 



473 



Plates under \ Inch in Thickness. 



Thickness of 
Plate. 
Inch. 


Width of Plate. 


Up to so 

Inches. 

Per Cent. 


50 Inches 

and Above. 

Per Cent. 


£ up to & 

A " "A 

A " " i 


IO 

7 


15 
IO 



Finish. 

15. Finished material must be free from injurious seams, flaws, or 
cracks, and have a workmanlike finish. 

Branding. 

16. Every finished piece of steel shall be stamped with the melt 
number, and steel for pins shall have the melt number stamped on the 
ends. Rivets and lacing steel, and small pieces for pin plates and 
stiff eners, may be shipped in bundles, securely wired together, with the 
melt number on a metal tag attached. 

Inspection. 

17. The inspector, representing the purchaser, shall have all reason- 
able facilities afforded to him by the manufacturer to satisfy him that 
the finished material is furnished in accordance with these specifica- 
tions. All tests and inspections shall be made at the place of manu- 
facture, prior to shipment. 

STEEL AXLES. 

Adopted 1901. Modified 1905. 

Process of Manufacture. 

1. Steel for axles shall be made by the open-hearth process. 

Chemical Properties. 

2. There shall be three classes of steel axles, which shall conform 
to the following limits in chemical composition : 



474 



APPLIED MECHANICS. 





Car and 

Tender-truck 

Axles. 

Per Cent. 


Driving and 
Engine -truck 

Axles. 

(Carbon Steel.) 

Per Cent. 


Driving-wheel 

Axles. 
(Nickel-steel.) . 

Per Cent. 


Phosphorus shall not exceed 

Sulphur " " " 

Manganese " " " 

Nickel 


0.06 
0.06 


O.06 
C.06 
0.60 


O.04 
O.04 

3 . to 4 . 





Physical Properties. 

3. For car and tender- truck axles, no tensile test shall 
be required. 

4. The minimum physical qualities required in the two classes of 
driving-wheel axles shall be as follows: 



Tensile Tests. 





Driving and 

Engine-truck 

Axles. 

(Carbon Steel.) 


Driving and 
Engine-truck 

Axles. 
(Nickel steel.) 




80,000 
40,000 

20 

25 


80,000 
50,000 

25 

45 




Elongation, per cent in two inches 


Contraction of area per cent 



5. One axle selected from each melt, when tested by the drop test 
described in paragraph No. 0, shall stand the number of 
rop es blows at the height specified in the following table without 
rupture and without exceeding, as the result of the first blow, the deflec- 
tion given. Any melt failing to meet these requirements will be rejected. 



Diameter of 


Number of 


Height of 


Deflection. 


Axle at Center, 




Drop. 




Inches. 




Feet 


Inches. 


4l 


5 


24 


H 


4f 


5 


26 


H 


4& 


5 


28J 


■ H 


4l 


5 


31 


8 


4f 


5 


34 


8 


5f 


5 


43 


7 


51 


7 


43 


5l 



6. Carbon-steel and nickel-steel driving-wheel axis shall not be 
subject to the above drop test. 



STEEL AXLES. 47 S 



Test Pieces and Methods of Testing. 

7. The standard test specimen one-half inch (J") diameter and 
two inch (2") gauged length shall be used to determine 

the physical properties specified in paragraph No. 4. It men fo^Ten- 
is shown in Fig. 1. (See p. 398.) 

8. For driving and engine- truck axles one longitudinal test specimen 
shall be cut from one axle of each melt. The center of Numberand 
this test specimen shall be half-way between the center j^f/ie 5 of •. 
and outside of the axle. mens - 

9. The points of supports on which the axle rests during tests must 
be three feet apart from center to center; the tup must DropTest 
weigh 1,640 pounds; the anvil, which is supported on Described. 
springs, must weigh 17,500 pounds; it must be free to move in a ver- 
tical direction; the springs upon which it rests must be twelve in number, 
of the kind described on drawing; and the radius of supports and of 
the striking face on the tup in the direction of the axis of the axle must 
be five (5) inches. When an axle is tested, it must be so placed in the 
machine that the tup will strike it midway between the ends; and it 
must be turned over after the first and third blows, and, when required, 
after the fifth blow. To measure the deflection after the first blow, 
prepare a straight edge as long as the axle, by reinforcing it on one side, 
equally at each end, so that, when it is laid on the axle, the reinforced 
parts will rest on the collars or ends of the axle, and the balance of the 
straight edge not touch the axle at any place. Next place the axle in 
position for test, lay the straight edge on it, and measure the distance 
from the straight edge to the axle at the middle point of the latter. 
Then, after the first blow, place the straight edge on the now bent axle 
in the same manner as before, and measure the distance from it to that 
side of the axle next to the straight edge at the point farthest away from 
the latter. The difference between the two measurements is the de- 
flection. The report of the drop test shall state the atmospheric tem- 
perature at the time the tests were made. 

10. The yield-point specified in paragraph No. 4 shall be determined 
by the careful observation of the drop of the beam or halt 

in the gauge of the testing machine. ie " poul ' 



476 



APPLIED MECHANICS. 



ii. Turnings from the tensile-test specimen of driving and engine- 
truck axles, or drillings taken midway between the center 
Sample for and outside of car, engine, and tender-truck axles, or 
Analysis. drillings from the small test ingot, if preferred by the 
inspector, shall be used to determine whether the melt is 
within the limits of chemical composition specified in paragraph No. 2. 

Finish. 

12. Axles shall conform in sizes, shapes, and limiting weights to the 
requirements given on the order or print sent with it. They shall be 
made and finished in a workmanlike manner, and shall be free from 
all injurious cracks, seams, or flaws. In centering, sixty- (60) degree 
centers must be used, with clearance given at the point to avoid dulling 
the shop lathe centers. 

Branding. 

13. Each axle shall be legibly stamped with the melt number and 
initials of the maker at the places marked on the print or indicated by 
the inspector. 

Inspection. 

14. The inspector, representing the purchaser, shall have all reason- 
able facilities afforded to him by the manufacturer to satisfy him that 
the finished material is furnished in accordance with these specifications. 
All tests and inspections shall be made at the place of manufacture, 
prior to shipment. 

STEEL TIRES. 
Adopted 1 90 1. 

Process of Manufacture. 

1. Steel for tires may be made by either the open-hearth or crucible 

process. 

Chemical Properties. 

2. There will be three classes of steel tires which shall conform 
to the following limits in chemical composition: 





Passenger 
Engines. 

Per Cent. 


Freight-engine 

and 

Car-wheels. 

Per Cent. 


Switching- 
engines. 

Per Cent. 




0.80 
0.20 
0.05 
O.05 


O.80 
O.20 
O.05 
0.05 


O.80 
O.20 

0.05 

0.05 


Silicon shall not be less than 

Phosphorus shall not exceed 

Sulphur shall not exceed 





STEEL 


TIRES. 




477 


Physical Properties. 
3. The minimum physical qualities required in each of 
the three classes of steel tires shall be as follows: 




Passenger- 
engines. 


Freight- 
engine and 
Car- wheels. 


Switching- 
engines. 


Tensile strength, pounds per square inch. 
Elongation, per cent in two inches 


100,000 
12 


110,000 
IO 


120,000 
8 



Drop Test. 



4. In the event of the contract calling for a drop test, a test tire 
from each melt will be furnished at the purchaser's expense, 
provided it meets the requirements. This test tire shall 
stand the drop test described in paragraph No. 7, without breaking or 
cracking, and shall show a minimum deflection equal to D 2 -f- 
(4oT 2 +2D), the letter "D" being internal diameter and the letter 
"T " thickness of tire at center of tread. 

Test Pieces and Methods of Testing. 

5. The standard turned test specimen, one-half inch (J") diameter 
and two inch (2") gauged length, shall be used to determine Test speci- 
the physical properties specified in paragraph No. 3. It Tensile Tests. 
is shown in Fig. 1. (See p. 398.) 

6. When the drop test is specified, this test specimen shall be cut cold 
from the tested tire at the point least affected by the drop Location of 
test. If the diameter of the tire is such that the whole men's. 6 P6C '~ 
circumference of the tire is seriously affected by the drop test, or if no 
drop test is required, the test specimen shall be forged from a test ingot 
cast when pouring the melt, the test ingot receiving, as nearly as pos- 
sible, the same proportion of reduction as the ingots from which the 
tires are made. 

7. The test tire shall be placed vertically under the drop in a run- 
ning position on solid foundation of at least ten tons in Dro Test 
weight and subjected to successive blows from a tup weigh- Described, 
ing 2,240 pounds, falling from increasing heights until the required 
deflection is obtained. 

8. Turnings from the tensile specimen, or drillings from the small 
test ingot, or turnings from the tire, if preferred by the 
inspector, shall be used to determine whether the melt is chemical* 
within the limits of chemical composition specified in naysis * 
paragraph No. 2. 



478 



APPLIED MECHANICS. 



Finish. 

9. All tires shall be free from cracks, flaws, or other injurious im- 
perfections, and shall conform to dimensions shown on drawings fur- 
nished by the purchaser. 

Branding. 

10. Tires shall be stamped with the maker's brand and number in 
such a manner that each individual tire may be identified. 

Inspection. 

11. The inspector representing the purchaser shall have all reason- 
able facilities afforded to him by the manufacturer to satisfy him that 
the finished material is furnished in accordance with these specifications. 
All tests and inspections shall be made at the place of manufacture, 
prior to shipment. 

STEEL RAILS. 



Adopted 1901. 
Process of Manufacture. 

1. (a) Steel may be made by the Bessemer or open-hearth process. 

(b) The entire process of manufacture and testing shall be in accord- 
ance with the best standard current practice, and special care shall be 
taken to conform to the following instructions: 

(c) Ingots shall be kept in a vertical position in pit heating furnaces. 

(d) No bled ingots shall be used. 

(e) Sufficient material shall be discarded from the top of the ingots 
to insure sound rails. 

Chemical Properties. 

2. Rails of the various weights per yard specified below shall con- 
form to the following limits in chemical composition: 



Carbon 

Phosphorus shall not 
exceed 

Silicon shall not ex- 
ceed 

Manganese 



50 to 59 + 
Pounds. 
Per Cent. 



0.35-0.45 J 

o. 10 

0.20 
o . 70-1 . 00 



60 to 60 + 

Pounds. 

Per Cent. 



O . 38-O . 48 

0. 10 

O.20 
o. 70-1.00 



70 to 79 + 

Pounds. 

Per Cent. 



o . 40-0 . 50 

0.10 

0.20 
0.75-1.05 



80 to 89 + 
Pounds. 
Per Cent. 



o- 43-o -53 
0.10 

0.20 
0.80-1. 10 



90 to 100 
Pounds. 
Per Cent. 



0.45-0.55 
0.10 



0.20 
0.80-1. 10 



STEEL RAILS. 



479 



Physical Properties. 
3. One drop test shall be made on a piece of rail not more than six 

Drop Test. 



feet long, selected from every fifth blow of steel. The rail 
shall be placed head upwards on the supports, and the 
various sections shall be subjected to the following impact tests: 



Weight of Rail. 
Pounds per Yard. 


Height of 
Drop. 
Feet. 


45 to and including 55 ... . 
More than 55 " " 65. . . . 

" 55 " " 75---- 

11 " 75 " " 8S--- 

" 85 " " 100.... 


15 
16 

17 
18 

19 



If any rail break when subjected to the drop test, two additional tests 
will be made of other rails from the same blow of steel, and, if either of 
these latter tests fail, all the rails of the blow which they represent will be 
rejected; but, if both of these additional test pieces meet the require- 
ments, all the' rails of the blow which they represent will be accepted. 
If the rails from the tested blow shall be rejected for failure to meet the 
requirements of the drop test, as above specified, two other rails will 
be subjected to the same tests, one from the blow next preceding, and 
one from the blow next succeeding the rejected blow. In case the 
first test taken from the preceding or succeeding blow shall fail, two 
additional tests shall be taken from the same blow of steel, the accept- 
ance or rejection of which shall also be determined as specified above; 
and, if the rails of the preceding or succeeding blow shall be rejected, 
similar tests may be taken from the previous or following blows, as the 
case may be, until the entire group of five blows is tested, if necessary. 

The acceptance or rejection of all the rails from any blow will 
depend upon the result of the tests thereof. 

Test Pieces and Methods of Testing. 
4. The drop-test machine shall have a tup of two thousand (2,000) 
pounds weight, the striking face of which shall have a Dro 
radius of not more than five inches (5")* and the test rail Machine. 
shall be placed head upwards on solid supports three test (3') apart. 
The anvil-block shall weigh at least twenty thousand (20,000) pounds, 
and the supports shall be a part of, or firmly secured to, the anvil. 



48O APPLIED MECHANICS. 

The report of the drop test shall state the atmospheric temperature at 
the time the tests were made. 

5. The manufacturer shall furnish the inspector daily with carbon 

determinations of each blow, and a complete chemical 
Chemical* analysis every twenty-four hours, representing the average 

of the other elements contained in the steel. These analy- 
ses shall be made on drillings taken from a small test ingot. 

Finish. 

6. Unless otherwise specified, the section of rail shall be the Amer- 

ican Standard, recommended by the American Society 
of Civil Engineers, and shall conform, as accurately as 
possible, to the templet furnished by the railroad company, consistent 
with paragraph No. 7, relative to specified weight. A variation in 
height of one-sixty-fourth of an inch (^VO less and one-thirty-second 
of an inch {$1") greater than the specified height will be permitted. A 
perfect fit of the splice-bars, however, shall be maintained at all times. 

7. The weight of the rails shall be maintained as nearly as possible, 

after complying with paragraph No. 6, to that specified in 
contract. A variation of one-half of one per cent (£%) for 

an entire order will be allowed. Rails shall be accepted and paid for 

according to actual weights. 

8. The standard length of rails shall be thirty feet (30'). Ten pe 

cent (10%) of the entire order will be accepted in shorter 
lengths, varying by even feet down to twenty-four feet 

(240- A variation of one-fourth of an inch (J") in length from that 

specified will be allowed. 

9. Circular holes for splice-bars shall be drilled in accordance with 

the specifications of the purchaser. The holes shall ac- 
curately conform to the drawing and dimensions furnished 
in every respect, and must be free from burrs. 

10. Rails shall be straightened while cold, smooth on head, sawed 

square at ends, and prior to shipment shall have the burr 
occasioned by the saw-cutting removed, and the ends made 

clean. No. 1 rails shall be free from injurious defects and flaws of all 

kinds. 

Branding. 

11. The name of the maker, the month and year of manufacture, 



STEEL SPLICE-BARS. 48 f 

shall be rolled in raised letters on the side of the web, and the number 
of the blow shall be stamped on each rail. 

Inspection. 

12. The inspector, representing the purchaser, shall have all reason- 
able facilities afforded to him by the manufacturer to satisfy him that 
the finished material is furnished in accordance with these specifications. 
All tests and inspections shall be made at the place of manufacture, 
prior to shipment. 

No. 2 Rails. 

13. Rails that possess any injurious physical defects, or which 
for any other cause are not suitable for first quality, or No. 1 rails, 
shall be considered as No. 2 rails, provided, however, that rails which 
contain any physical defects which seriously impair their strength shall 
be rejected. The ends of all No. 2 rails shall be painted in order to 
distinguish them. 

STEEL SPLICE-BARS. 

Adopted 1 90 1. 

Process of Manufacture. 

1. Steel for splice-bars may be made by the Bessemer or open- 
hearth process. 

Chemical Properties. 

2. Steel for splice-bars shall conform to the following limits in 
chemical composition: 

Per Cent. 

Carbon shall not exceed o . 15 

Phosphorus shall not exceed o . 10 

Manganese o . 30-0 . 60 

Physical Properties. 

x. Splice-bar steel shall conform to the following physi- 
, f. . ° r J Tensile Tests. 

cal qualities: 

Tensile strength, pounds per square inch 54>°°° to 64,000 

Yield-point, pounds per square inch 32,000 

Elongation, per cent in eight inches shall not be 

less than 25 



482 APPLIED MECHANICS. 

4. (a) A test specimen cut from the head of the splice-bar shall 
_ .. bend 180 flat on itself without fracture on the outside of 

Bending 

Tests. the bent portion. 

(b) If preferred, the bending tests may be made on an unpunched 
splice-bar, which, if necessary, shall be first flattened, and shall then be 
bent 180 flat on itself without fracture on the outside of the bent por- 
tion. 

Test Pieces and Methods of Testing. 

Test sped- 5- A test specimen of eight inch (8") gauged length, cut 

Te e nsne r Tests. ^ rom tne nea ^ of the splice-bar, shall be used to determine 
the physical properties specified in paragraph No. 3. 

6. One tensile-test specimen shall be taken from the rolled splice- 

bars of each blow or melt; but in case this develops flaws, 
Number of or breaks outside of the middle third of its gauged length, 
Tensile Tests. ^ ma y k e discarded, and another test specimen substituted 

therefor. 

7. One test specimen cut from the head of the splice-bar shall be 

taken from a rolled bar of each blow or melt, or, if preferred, 
men for*'" the bending test may be made on an unpunched splice-bar 
Bending. which, if necessary, shall be flattened before testing. The 
bending test may be made by pressure or by blows. 

8. For the purposes of this specification the yield-point shall be de- 

termined by the careful observation of the drop of the beam 
Yield-point. 0f halt in the gauge of the test ; ng mac hine. 

9. In order to determine if the material conforms to the 
Sample for chemical limitations prescribed in paragraph No. 2 herein, 
Anafrsfe! analysis shall be made of drillings taken from a small test 
ingot. 

Finish. 

10. All splice-bars shall be smoothly rolled and true to templet. 
The bars shall be sheared accurately to length and free from fins and 
cracks, and shall perfectly fit the rails for which they are intended. The 
punching and notching shall accurately conform in every respect to the 
drawing and dimensions furnished. A variation in weight of more 
than 2\ per cent from that specified will be sufficient cause for rejection. 



STRENGTH OF STEEL. 483 

Branding. 

11. The name of the maker and the year of manufacture shall be 
rolled in raised letters on the side of the splice-bar. 

Inspection. 

12. The inspector, representing the purchaser, shall have all reason- 
able facilities afforded to him by the manufacturer, to satisfy him that 
the finished material is furnished in accordance with these specifications. 
All tests and inspections shall be made at the place of manufacture, 
prior to shipment. 

§ 226. Strength of Steel. — The literature upon steel is 
exceedingly voluminous, and many books and articles written 
upon the metallurgy of steel, such as "Metallurgy of Steel," by 
Henry M. Howe, and "The Manufacture and Properties of Iron 
and Steel," by H. H. Campbell, contain a great many tests, which 
have, as a rule, to do with its properties and the effects of different 
compositions and treatments. They do not often contain; how- 
ever, tests upon full-size pieces, such as columns for bridges or 
buildings, beams, large riveted joints, full-size parts of machinery, 
etc. The greater part of this latter class of tests are to be found in 
the reports of the various testing laboratories, such as those of 
the laboratories at Munich, at Berlin, and at Zurich in Europe, 
and the Watertown Arsenal reports and the Technology Quarterly 
in America; and also in various articles in the Proceedings of 
the various Engineering Societies in Europe and America. A 
number of these have already been mentioned among the refer- 
ences to tests of wrought-iron, and the greater part of them contain 
also experiments on steel. 

References to such full-size tests of steel as are quoted here 
will be given in connection with the tests themselves. 

A detailed study of the effect of the different ingredients 
and combinations of ingredients, upon the strength, elasticity, 
and ductility of steel, is a very complicated matter; it belongs 
to the study of Metallurgy and is beyond the scope of this 
work. Nevertheless, the engineer needs, of course, some general. 



4^4 APPLIED MECHANICS. 



knowledge of these matters, and especially of the effect, within 
certain limits, of different percentages of carbon. 

This subject has been dealt with by Mr. Wm. R. Webster 
in the Trans. Am. Inst. Mining Engineers, of October, 1892, 
August, 1893, and October, 1898, and in the Journal of the Iron 
and Steel Institute, No. 1, 1894; also by Mr. A. C. Cunningham 
in the Trans. Am. Soc. Civil Engineers of December, 1897; and 
by Mr. H. H. Campbell, in his book, " Metallurgy of Iron and 
Steel." Of course none of them claims anything more than 
approximation for their various rules and formulae, and then only 
in the case of what they call normal steel, i.e., such steel as is 
most frequently manufactured by the mills. 

Mr. Webster made an investigation of the effects of carbon, 
phosphorus, manganese, and sulphur upon the tensile strength 
of the steel. He gives a set of tables from which to determine, 
approximately, the tensile strength of normal steel, of a given 
chemical composition. His investigations were principally made 
upon basic Bessemer, and basic open-hearth steel. 

Mr. Campbell gives a formula for the tensile strength of acid, 
and another for the tensile strength of basic steel, and states that 
they represent the facts with a good degree of accuracy. His 
formulae are as follows : 

For acid steel, 

38600 + 1 2 1 C + 89P + R = ultimate strength ; 

For basic steel, 

3743o + 95C + 8.5Mn + io5P + R = ultimate strength; 

where C indicates carbon, P phosphorus, and Mn manganese, 
in units of 0.001 per cent, and R depends upon the finishing 
temperature, and may be plus or minus. 

Mr Cunningham gives the following rule: To find the approx- 
imate tensile strength of structural steel; to a base of 40000 add 
1000 pounds for every 0.01 per cent of carbon, and 1000 pounds for 



STRENGTH OF STEEL. 



485 



every 0.01 per cent of phosphorus, neglecting all other elements 
in normal steel. 

In this connection a set of tests will be quoted which were 
made on the government testing-machine at Watertown Arsenal, 
upon specimens of steel containing different percentages of carbon, 
the tests themselves forming a portion of a series denominated 
in the government report as the "Temperature Series." The 
account of the tests to-be quoted is to be found in their report 
for 1887. 

Ten grades of open-hearth steel are here represented, in which 
the carbon ranges from 0.09 to 0.97 per cent, varying by tenths 
of a per cent as nearly as was practicable to obtain the steel. 

The other elements do not follow any regular succession. 



TENSILE TESTS OF STEEL BARS— TEMPERATURE SERIES. 
Tests at Atmospheric Temperature. 



r 










=• 


u 


' 


ad 




a 


a> 1) 


rt.S 


rtJ3 


c 














a 






ac 




2-cu 






O 

H 

C 


« 
U 

u 
1) 

- 

a 




U 

u 

- 

8 



c 

bo 

G 

CS 


c 
u 



u 

V 

- 

c 




en 

u 



c 

a 

CO 


cr 
a 

< 

"3 

c 
,o 



V 

in 


CO 

.0 
•J 

vO~ 
00 
00 

C = 

Is 


0] 

V 

<* 

3 a 

txc 
c § 


& 
OC/3 

JS a 


is ©.J 


a 
§£ 

C V 


Is 

t«u 


11. 

rt tec 

.C rt u 

8W = 


echanical Work 
Tensile Streng 
in Inch-Lbs. 


V V 


» 


u 


s 


Ui 


p 


J 


1-1 


M 


h 


w 


u 


§ 


£ 


(2 


753 


0.09 


O.II 




1.009 


0.80 


21000 


3 


30000 


52475 


23.6 


63-5 


I5-85 


9808.36 


106434 


754 


0.20 


0.45 




1.009 


0.80 


25000 


3 


395oo 


68375 


21.2 


49.1 


26.40 


10651.90 


"3704 


755 


0.31 


057 




0.798 


0.50 


25000 


6 


46500 


80600 


18.O 


43-5 


37-27 


10660.77 


126640 


756 


0.37 


0.70 




0.798 


0.50 


25000 


6 


50000 


85160 


17. 5 


45-3 


42.50 


10935-48 


134600 


757 


0.51 


0.58 


0.02 


0.798 


0.50 


30000 


6 


58000 


98760 


14.9 


41.6 


58.00 


11380.62 


152380 


758 


o-57 


093 


0.07 


0.798 


0.50 


30000 


6 


55000 


I I 7440 


10. 1 


14.0 


52-43 


11169.34 


134880 


759 


0.71 


0.58 


0.08 


o-757 


0.45 


35000 


12 


57000 


116000 


8.8 


26.2 


56.53 


9231.21 


151510 


7 6o 


0.81 


0.56 


0.17 


0.798 


0.50 


40000 


12 


70000 


149600 


5-o 


5-4 


84-35 


7872.20 


158140 


761 


0.89 


0.57 


0.19 


o-757 


o-45 


45000 


12 


75000 


141290 


4-3 


4-4 


95.00 


6418.53 


147860 


762 


0.97 


0.80 


0.28 


Q-757 


o-45 


50000 


12 


79000 


152550 


4-3 


5-8 


108.62 


7550.23 


161910 
























— 



486 



APPLIED MECHANICS. 



The following tables include sets of miscellaneous tests of 
various kinds of steel. 



Bessemer Steel. 


Open-hearth Steel. 


5 c 

II 


1 Maximum 
I Load. 

(Lbs. per 
1 sq. in.) 


1 Elastic 
1 Limit. 
1 (Lbs. per 
1 sq. in.) 


0^ 

.„ re aj 

-jJJ u 

3 <! >- 


II 


3 . 

V TO 

n 
7600 


a s 

3 a—. 

64169 


4) 
— a ">•- 

5jJ cr 
47395 


c . « 
a c 

tS£8 
56.7 


in u 

3\r; 

3JS 


.7426 


70983 


40397 


54-7 


29139000 


29392600 


.7481 


57760 


33000 


53-5 


28885000 


7500 


63083 


44137 


64.0 


30179000 


• 74 6 3 


58408 


28575 


58.9 


32799000 


7600 


64.477 


47394 


60.1 


30780000 


.7285 


62761 


34787 


65.2 


32135000 


7700 


62449 


46171 


63.5 


30481000 


.7476 


50505 


19364 


72-5 


29479000 


7700 


62556 


46171 


64.3 


29073000 


.7442 


51230 


19550 


69.7 


30653000 


7700 


62857 


46171 


59-5 


29073000 


.7500 


51110 


21503 


7i-5 


28457000 


7600 


643 '5 


45189 


64. i 


29843000 


.7500 


51518 


21503 


5 °/l 


27665000 


7650 


63527 


44600 


64.6 


29008000 


.7300 


73865 


46584 


56.8 


29600000 


7600 


64830 


42984 


61.8 


28527000 


.7500 


50294 


26029 


27.0 


28055000 


7550 


65020 • 


45790 


57-9 


29338000 


.7400 


97655 


54641 


44.8 


30539000 


7600 


65140 


45190 


64.9 


31288000 


.7400 


87086 


47666 


46.8 


30090000 


7575 


65240 


43270 


62.3 


3004000Q 


•735o 


87508 


50673 


48.0 


30057000 


7575 


65125 


45487 


59 9 


29912000 


.7600 


65235 


49598 


62.5 


30310000 


7600 


64500 


40780 


61.9 


30060000 


•735o 


87014 


50673 


45-o 


30058000 


7550 


650S9 


41320 


61.3 


291340,0 


.7400 


87356 


4999i 


38.6 


30090000 


.87 


43300 


21900 


75.6 


28500000 


.7500 


84317 


48665 


46.2 


28868000 


•73 


44900 


21500 


75-7 


29900000 


.74 20 


86720 


49650 


47.2 


29887000 


•72 


46300 


22100 


73-6 


29900000 


. .7691 


66465 


35526 


61.5 


29149000 


.60 


46800 


24800 


73-3 


30800000 


• 773o 


66077 


35159 


61.5 


30244000 


.60 


46700 


24800 


75-o 


30000000 


.7690 


66745 


35526 


62.9 


30075000 










.7690 


66445 


39832 


62.5 


30560000 










.7690 


66142 


35526 


60.7 


27864000 










.7680 


66530 


3S 6i8 


60.9 


29225000 










.7690 


67068 


39832 


61.5 


30075000 











Machine-steel. 


<u 


6 S3 




rt c 


o£ 


' -0-4J 

.2 a 


3 CL~ 


tic 
lit. 
s. p 
in.) 


3< v. 


!'•- 


Z. rt 


X g.O . 


2 B -° • 


•a « 


•O c« 


.«8 S 




JS-j J cr 


f*°~ 


S S 


.7608 


91795 


62693 


53-3 


293160OO 




7629 


96256 


66723 


44.2 


2939IOOO 




7633 


96767 


65561 


44.1 


29586000 




7520 


92087 


59665 


40.4 


30482OOO 




7593 


92091 


62940 


54-o 


308480OO 




7598 


92191 


58445 


50.4 


2896800O 




7625 


86941 


62413 


48.7 


2834OOOO 




7623 


95750 


62445 


46.2 


306040OO 




7620 


06045 


62495 


44.6 


288020OO 




7620 


91220 


62495 


5T-6 


3270600O 




7560 


96684 


63490 


42.8 


294OOOOO 




7634 


96567 


66638 


43-3 


3051800O 




7609 


92804 


60755 


5i-3 


28884OOO 




7597 


86678 


58460 


50.2 


29867OOO 




7580 


96119 


54292 


41.8 


29818000 




7600 


86741 


58415 


45-7 


28738000 




75^3 


99635 


62032 


27.4 


218130OO 




7613 


106980 


60413 


48.4 


292910OO 




7590 


96142 


54 J 49 


45-6 


26643OOO 




7699 


94513 


59071 


46.6 


281480OO 




7622 


91435 


64654 


54-2 


271640OO 




7613 


86775 


60413 


47.8 


2929IOOO 




7567 


96249 


61 1 50 


45-2 


287760OO 




7579 


95303 


67606 


44.9 


30OO500O 




754° 


95518 


54870 


40.8 


29416000 




7554 


84990 


45752 


55-5 


2975IOOO 



Boiler-plate. 



Section. 



379 xi 
384 x 1 
365 XI 
369 xi 
398 xi 
376 xi 
4095 x 1 
375 xi 
3737X1 
475 x 1 
4702 x 1 
4292 X I 
4258 X I 
4225 X I 

4125x1 

40 X I 
50 X I 

49 x 1 
49x1 

50 X I 

49 x 1 

50 x 1 
49 x 1 
39 x 1 
38 x 1 

41 XT 



.458 
.48 
.65 
•49 

•5" 

.496 

•3647 

•494 

•4974 

•0295 

.0064 

.0235 

.0123 

.0025 

.0025 

.02 

.02 






59450 
58770 
58115 
61657 
54370 
53156 
54930 
55173 
54220 
47954 
51035 
55560 
54984 
60680 
61500 
57190 
60352 
63825 
60024 
59803 
61024 
60393 
63625 
50480 
53543 
58144 



Im 

u 

. 00 


c *_; 

•2 ** £ - 


►. 

5 -8 


y.t: • c 


O u O 




w e.o . 


3< u 


•a rt 


JS"j J cr 


2*t 


sS 











31670 


47-3 


2Q45900O 


39590 


56.3 


30270000 


32380 


45-6 


29305000 


37284 


5i-9 


30135000 


30760 


58.0 


28608000 


32889 


57-1 


28511000 


33"o 




29826000 


29451 


58.5 


28849000 


31270 


52.5 
673 
67.5 
64 -9 
67.7 

56-3 
55-3 


27490000 






26552 


58.5 


25800000 


28431 


58.8 


36109000 


31010 


52.0 


25273000 


29012 


58.8 


26677000 


29012 


50.5 


30012000 


29012 


58.8 


30012000 


29412 


60.2 


29412000 


30012 


46.1 


3 1 866000 


26041 


62.5 


32051000 


29009 


59-8 


26168000 


27846 


50.1 


35455000 



TENSILE STRENGTH OF STEEL. 



487 



Bessemer Steel Wire. 



Diameter 
of Cross- 
section. 
(Inches.) 



, 1290 
1280 
,1288 
,T2yr 
,1283 
1283 
1289 
■128 1 
1283 
1286 



Elastic 

Limit. 

(Lbs. per 

sq. in.) 



66100 
68300 
67200 
66500 
69600 
69700 
71400 
65800 
68500 



Maximum 

Load. 
(Pounds.) 



1013 
1021 
1010 
970 
996 
102 1 
1005 
1043 
1004 
1000 



Maximum 

Load. 
(Lbs. per 

sq. in.) 



77500 
79400 
77500 
74100 
77000 
79000 
77000 
80900 
77700 
77000 



Reduction 

of Area. 

(Per cent.) 



Bessemer Spring-steel Wire. 



Modulus of 

Elasticity. 

(Lbs, per 

sq. in. 



.0911 


75000 


950 


146000 


34-6 


.0910 


69900 


974 


149000 


Si- 6 


.0905 


79600 


969 


150000 


42.0 


.0911 


72900 


950 


146000 


37-5 


.0905 




931 


143000 


39-6 



30000000 
30400000 
30900000 
30000000 
28500000 
30000000 
31200000 
29200000 
30700000 
31000000 



24500000 
25900000 
23000000 
24200000 
25400000 



Tests of Steel Eye-bars. 

Tests of Steel Eye- bars made on the Government Machine. 
— In the Tests of Metals at Watertown Arsenal for 1883 is 
the record of the tests of six eye-bars of steel, presented by 
the president of the Keystone Bridge Company. 

The following is an extract from the report in regard to 
these eye-bars: 

" The eye-bars were made of Pernot open-hearth steel, fur- 
nished by the Cambria Iron Company of Johnstown, Penn. 

"The furnace charges, about 15 tons each of cast-iron, 
magnetic ore, spiegeleisen, and rail-ends, preheated in an aux- 
iliary furnace, required six and one-half hours for conversion. 

" All these bars were rolled from the same ingot. 

" Samples were tested at the steel-works taken from a test 
ingot about one inch square, from which were rolled |-inch 
round specimens. 



488 



APPLIED MECHANICS. 



"The annealed specimen was buried in hot ashes while still 
red-hot, and allowed to cool with them. 

"The following results were obtained by tensile tests : — 



f-inch round rolled bar . 

f-inch round rolled and 

annealed bar .... 



Elastic 
Limit, in 
lbs., per 

Sq. In. 



4804O 
42210 



Ultimate 

Strength, ii 

lbs., per 

Sq. In. 



73150 
69470 



Contrac- 
tion of 
Area. 



%■ 

457 
54-2 



Modulus 

of 
Elasticity. 



282IOOOO 
292IOOOO 



Carbon. 



O.27 
O.27 



"The billets measured 7 inches by 8 inches, and were 
bloomed down from 14-inch square ingot. 

"They were rolled down to bar-section in grooved rolls at 
the Union Iron Mills, Pittsburgh. 

" The reduction in the roughing-rolls was from 7 inches by 
8 inches to 6| inches by 4 inches ; and in the finishing-rolls, to 
6\ inches by 1 inch. 

"The eye-bar heads were made by the Keystone Bridge 
Company, Pittsburgh, by upsetting and hammering, proceeding 
as follows : — 

"The bar is heated bright red for a length of (approxi- 
mately) 27 inches, and upset in a hydraulic machine ; after 
which the bar is reheated, and drawn down to the required 
thickness, and given its proper form in a hammer-die. 

" The bars are next annealed, which is done in a gas-furnace 
longer than the bars. They are placed on edge on a car in the 
annealing-furnace, separated one from another to allow free 
circulation of the heated gases. They are heated to a red heat, 
when the fires are drawn, and the furnace allowed to cool. 
Three or four days, according to conditions, are required before 
the bars are withdrawn. 



TENSILE STRENGTH OF STEEL. 489 

" The pin-holes are then bored. 

"The analyses of the heads before annealing were: — 

" Carbon, by color 0.270 per cent. 

Silicon 0.036 " 

Sulphur * 0.075 " 

Phosphorus 0.090 " 

Manganese 0.380 " 

Copper . Trace. 

" The bars were tested in a horizontal position, secured at 
the ends, which were vertical. 

" To prevent sagging of the stem, a counterweight was used 
at the middle of the bar. 

" Before placing in the testing-machine, the stem from neck 
to neck was laid off into 10-inch sections, to determine the 
uniformity of the stretch after the bar had been fractured. 

"A number of intermediate 10-inch sections were used as 
the gauged length, obtaining micrometer measurements of 
elongation, and the elastic limit for that part of the stem which 
was not acted upon during the formation of the heads. Elon- 
gations were also measured from centre to centre of pins, taken 
with an ordinary graduated steel scale. 

" The moduli of elasticity were computed from elongations 
taken between loads of 10000 and 30000 lbs. per square inch, 
deducting the permanent sets. 

"The behavior of bars Nos. 4582 and 4583, after having 
been strained beyond the elastic limit, is shown by elongations 
of the gauged length measured after loads of 40000 and 50000 
lbs. per square inch had been applied ; and with bar No. 4583, 
after its first fracture under 64000 lbs. per square inch, a rest 
of five days intervening between the time of fracture and the 
time of measuring the elongations. 

" Considering the behavior between loads of 10000 and 
30000 lbs. per square inch, we observe the elongations for the 



49° APPLIED MECHANICS. 

. » 

primitive readings are nearly in exact proportion to the incre- 
ments of load. 

" Loads were increased to 40000 lbs. per square inch, passing 
the elastic limit at about 37000 lbs. per square inch ; the respec- 
tive permanent stretch of the bars being 1.3 1 and 1.26 per cent. 

"Elongations were then immecfiately redetermined, which 
show a reduction in the modulus of elasticity, as we advanced 
with each increment, of 5000 lbs. per square inch. 

"Corresponding measurements after the bars had been 
loaded with 50000 lbs. per square inch reach the same kind of 
results. 

"The first fracture of bar No. 4583, under 64000 lbs. per 
square inch, occurred at the neck, leaving sufficient length to 
grasp in the hydraulic jaws of the testing-machine, and con- 
tinue observations on the original gauged length. This was 
done after the fractured bar had rested five days. 

"The elongations now show the modulus of elasticity con- 
stant or nearly so, the only difference in measurements being 
in the last figures, up to 50000 lbs. The readings were then 
immediately repeated, and the same uniformity of elongations 
obtained. 

" An illustration of the serious influence of defective metal 
in the heads is found in the first fracture of bar No. 4583. 

" There was about 27 per cent excess of metal along the 
line of fracture over the section of the stem." 






TENSILE STRENGTH OF STEEL. 



49 1 



8* 8* 8 s 



Gauged Length, in inches. 



Width, in inches. 



% 



3 H 



2 5= 3' 8 



;|S 



$ § r 



4 Cfl 



In Gauged 
Length. 



Centre to Cen- 
tre of Pins. 



s r 



Contraction of Area, per 
cent. 



log 



g^ 



Maximum Compression 
on Pin-Holes, in lbs., 
per Square Inch. 



I 

3' 

n 
3 



W td 
3. 3. 



t« 3 (« 






b, S a, 



o ►-: hi 



O Sfl 



492 



APPLIED MECHANICS. 



ELONGATIONS OF No. 4582 FOR EACH INCREMENT OF 5000 LBS. PER 

SQUARE INCH. 







Elongations. 




Loads, in lbs., 

per 
Square Inch. 








Primitive Load- 
ing. 


After Load of 
40000 lbs. per 
Square Inch. 


After Load of 
50000 lbs. per 
Square Inch. 


I OOOO 


_ 


. 




15000 


O.0274 


O.030O 


O.O31 1 


20000 


O.0269 


O.0305 


O.0322 


2500O 


O.0269 


O.O320 


0.0337 


30000 


O.O269 


O.O330 


O.O34I 



ELONGATIONS OF No. 4583 FOR EACH INCREMENT OF 5000 LBS. PER 

SQUARE INCH. 



Loads, in 

lbs., per 

Square Inch. 


Elongations. 


Elongations after 64000 lbs. per 
Square Inch. 


Primitive Load- 
ing. 


After 40000 lbs. 

per 

Square Inch. 


After 50000 lbs. 

per 

Square Inch. 


First 
Reading. 


Second 
Reading. 


I OOOO 
15000 
20000 
25000 
30OOO 
35000 
40OOO 
45000 
50000 


O.0272 
O.0272 
0.0268 
O.O267 


O.029I 
O.0305 
O.O314 
O.O326 


O.0302 
O.0315 
O.0325 
O.0340 


O.03 1 1 
O.0308 
O.031 1 
O.0312 
O.03 1 1 
O.0312 
O.0310 
O.0315 


O.0310 
O.0310 
O.031O 
O.0310 



In the Tests of Metals for 1886 is given the following table 
of tensile tests of steel eye-bars, furnished by the Chief Engineer 
of the Statue of Liberty. 



TENSILE STRENGTH OF STEEL. 



493 



Dimensions. 


u 


u 


Elongation. 


cS 




x. » 




Fracture. 




Pi 


■5 • 






1 










O i 














Length, Cente 
Center of P 
holes. 


i 

is 


09 

1 

O 


£ 


r 




t-iQ. 

3 


O 

.2 



u 

O 
O 


3 
ft 


Maximum Co 
sion on Pi 
per Square 


i 

1 




Appearance. 


Ins. 


Ins. 


Ins. 


Lbs. 


Lbs. 


% 


% 


% 


Lbs. 


Lbs. 






308.00 


S.16 


1 .02 


34610 


64870 


7-4 


7-3 




31400000 


74173 






308.00 


5-14 


1 .02 


347 3o 


69330 


10.4 


10.3 




29279000 


84093 






308.00 


5-15 


1 .02 


3733o 


70286 


n. 7 


11. 5 




29017000 


80043 






308. 10 


5-14 


1 .02 


35000 


70229 


11. 6 


11. 4 


13.4 




79826 


Stem 


Granular, radi- 
ating from a 
button of 
hard metal. 


308.00 
3°7-9S 


5-13 
5-15 




3595o 
35000 


71680 
70895 




11. 5 
11. 8 






81323 

80737 






1 .02 


.... 




30162000 







The gauged length of the bars was 260 inches. The moduli 
of elasticity computed between 25000 and 30000 pounds per 
square inch. 

In connexion with the work upon the bridge over the Missis- 
sippi at Memphis, Mr. Geo. S. Morison, the Chief Engineer, 
had 56 full-size stee eye-bars tested. The results are given in 
his Report, dated March, 1894, and furnish valuable information 
regarding the behavior of the steel, and the design, and cDn- 
struction of the bars. Only the following table (see page 494) 
will be given here, containing a portion of the results of the tests 
upon 31 of the bars, all made of basic open-hearth steel, and 
all of which broke in the body. 

This table will aid the reader in comparing the tensile strength 
and the limit of elasticity of full-size steel eye-bars, with those 
obtained from the tests of small samples of the steel. 

In Engineering News of Feb. 2, 1905, is an article containing 
a comparison of full-size and specimen tests of eleven steel eye- 
bars, made at the Phoenix Iron Co. Each of these bars was 15 
inches wide; two of them were i\ inches thick; one was i& 
inches thick, six were 2 inches thick, and two were 2ts inches 
thick. The specimen tests gave tensile strengths varying from 
60310 to 67000 pounds per square inch, and limits of elasticity 
varying from 31550 to 41760 pounds per square inch. 



494 



APPLIED MECHANICS. 







Full-size Eye-bars. 










u 












0) OJ 


j & 


I-* 




% 

is 


0> 

1 



'& 

Eh 


J* 


w 

03 Pi 
8 


.5 1-1 

as 

3 




Ins. 


Ins. 


Ins. 


Lbs. 


Lbs. 


I 


IO.07 


1 50 


160.63 


35i°° 


67490 


3 


9-95 


1 


73 


358.93 


37680 


70160 


4 


9.98 


1 


75 


361.23 


39700 


65500 


5 


10.05 


1 


5o 


162.38 


33 J 40 


65060 


6 


6.08 


1 


13 


291 .26 


29690 


56700 


8 


10.07 


1 


67 


287.37 


32860 


65600 


9 


9.92 


1 


95 


284.28 


31110 


61060 


IO 


9.94 





99 


287.88 


33990 


63220 


ii 


10.05 


2 


20 


222.88 


29330 


63 1 CO 


13 


10.12 


1 


86 


464.03 


31970 


53860 


i5 


7.12 


1 


17 


314.O4 


30270 


51500 


16 


10.07 


2 


20 


338.73 


28080 


55160 


18 


10.03 


I 


81 


25L58 


29670 


62140 


21 


9-97 


I 


37 


250.28 


32700 


65 400 


23 


7.02 


1 


3i 


385.73 


28980 


52010 


24 


7.01 


1 


26 


385.78 


28410 


54740 


25 


9.99 


1 


62 


249.98 


30500 


58870 


27 


9.96 


2 


05 


341.28 


3336o 


73550 


28 


10.13 


1 


30 


249.48 


32520 


60710 


30 


9.98 


1 


81 


284.82 


28000 


58720 


31 


i°-i5 


1 


83 


221.98 


32290 


62270 


34 


10.04 





99 


361.68 


29970 


58680 


35 


7.01 


1 


27 


258.68 


28640 


56830 


42 


7.98 


1 


20 


254.63 


3i93o 


63870 


43 


8.03 


2 


32 


338.58 


32840 


62400 


44 


7.00 


1 


18 


258.68 


27870 


53520 


46 


9.09 


1 


25 


206.58 


32590 


57410 


53 


8. 11 


1 


79 


279.98 


28940 


58010 


55 


7.00 


1. OO 


289.23 


31380 


59850 



Sample Bars from 


same Melt. 


1 


a 


¥. 


t3 

03 

O . 

.J a 
.5 ^ 


IN 


13 
g 


33 

5Ph 


►3 C7 

'-2 i- 


15 




c 


GO V 

<A ft 









W 


W 


S 


Ph 


Sq. In. 




Lbs. 


Lbs. 




.9500 


27-5 


41580 


73050 


.027 


.9918 


24.4 


42650 


75620 


.015 


•9520 


28.8 


40280 


70280 


.062 


.9500 


27.5 


41580 


73050 


.027 


•9756 


28.1 


40490 


69700 


.026 


I. 1590 


20.0 


43750 


75000 


.021 


I .0140 


28.8 


42210 


69730 


.046 


.9868 


28.1 


40230 


69720 


• 025 


•9635 


28.8 


38090 


71300 


.017 


I .0201 


27.0 


40200 


71860 


.017 


I. Ol8o 


28.8 


33400 


57i7o 


.014 


I. I220 


24.2 


38320 


70220 


•023 


1.0200 


26.3 


40200 


71080 


.028 


I .0670 


25.0 


3936o 


69360 


.041 


I . 1 700 


3*-3 


34190 


58460 


•039 


I. 0170 


28.1 


41400 


67840 


.OIO 


•9338 


25.0 


40910 


70360 


.014 


.9700 


25-5 


40410 


69900 


•063 


•9504 


27.0 


40400 


70490 


•023 


•5557 


29-5 


40000 


66800 


.008 


.9746 


21.3 


40530 


72240 


.056 


1. 1720 


27.0 


40610 


70480 


.060 


1 .0200 


28.1 


40790 


68730 


.030 


1. 0100 


21.9 


40900 


69800 


.024 


1 .0620 


23.1 


41710 


71000 


.066 


1.0560 


3i-9 


32480 


58050 


.027 


•9734 


28.7 


38110 


60920 


.014 


1. 114 


23.0 


40480 


66880 


.030 


1 .020 


28.1 


40790 


68730 


030 



The decrease of tensile strength in the full-size eye-bars 
varied from 6.3 per cent to 11.9 per cent, while the decrease in 
elastic limit varied from 8.3 per cent to 17 per cent. 

STEEL COLUMNS. 

In the Trans. Am. Soc. C. E., of June, 1889, will be found a 
paper by Mr. J. G. Dagron, giving an account of a set of tests of 
eight full-size Bessemer-steel bridge columns, made for the Sus- 



STEEL COLUMNS. 



495 



quehanna River Bridge of the Baltimore and Ohio R.R. The steel 
varied in tensile strength from 83680 to 84440 pounds per square 
inch, in elastic limit from 51 190 to 53890 pounds per square inch, 
in elongation in 8 inches from 18.75 per cent to 20.75 P er cent > 
and in contraction of area from 30.55 per cent to 39.7 per cent. 
The columns were made by the Keystone Bridge Company and 
tested in their hydraulic press, with the columns in a horizontal 
position, and with the pins horizontal. 

The results obtained are given by the accompanying table : 



No. of 


Depth. 


Sectional 
Area. 


Length 
Center to 


Ratio of 
Length to 


Square of 
Radius of 


Ultimate 

Strength, 

in Lbs. 


Modulus of 

Elasticitv. 

Lbs. 






Sq. Ins. 


Pin-holes. 


Gyration. 


Gyration. 


Sq. In. 


per Sq. In. 


1 


8 


8.24 


16' o" 


42 /05 


20.86 


41020 


27705000 


2 


8 


8 


24 


16' O" 


42 


o.S 


20.86 


41650 


27705000 


3 


8 


8 


24 


20' o" 


52 


564 


20.86 


39440 


2 6l I 3COO 


4 


8 


8 


24 


20' 


52 


5^4 


20.86 


41050 


2581 60CO 


5 


8 


8 


24 


2 4'o" 


63 


07.S 


20.86 


40230 


29504000 


6 


8 


8 


24 


2 4 'o" 


63 


o75 


20.86 


40070 


283980OO 


7 


9 


13 


23 


25'7-r 


5« 


795 


2 7-34 


35570 


26557OOO 


S 


9 


13 


23 


25-7-r 


58 


795 


27-34 


38810 


294780OO 



The columns failed as follows 



No. 



1. 



Failed by bending downwards at rivet in latticing, 1 foot 

\o\ inches from the center, buckling flange angles and 

web-plate. 
No. 2. Failed by bending upwards at rivet in latticing at center, 

buckling flange angles and web-plate. One angle 

was fractured at point of buckling, and also at the two 

adjacent rivets in latticing 
No. 3. Failed by bending upwards between latticing, 3 feet from 

center, buckling flange angles and web-plate. 
No. 4. Failed by bending upwards between latticing, 4 inches from 

center, buckling flange angles and web-plate. 
No. 5. Failed by bending upwards between latticing, 9 J inches 

from center, buckling flange angles and web-plate. 



496 



APPLIED MECHANICS. 



No. 6. Failed by bending upwards between 



latticing, i 



foot 



5! inches from center, buckling flange angles and web 

plate. 
No. 7. Failed by bending upwards at rivet in latticing, 3 inches 

from center, buckling flange angles and web-plate. 
No. 8. Failed by bending upwards at rivet in latticing, 1 foot 

from center, buckling flange angles and web-plate. 

In every case, after test, the rivets of each column were found 
by hammer test to be perfectly right. 

The following table gives the results of a set of tests by direct 
compression, of eight connecting-rods specially made for these 
tests, by the Baldwin Locomotive Works, and tested in the Labora- 
tory of Applied Mechanics of the Mass. Institute of Technology. 













Breaking- 








Area. 


Tensile Properties of the Steel. 


strength per Sq. 




S 


I 






In. of the Rod. 





c 



43 




CO 





Modulus 





c* 








P 







W 




U«3 


s 

O w 

c5^j 


.2 


of 
Elasticity 




to 




03 

CO 





cu2 




4J 


■g 

3 


+3 «H 

rt ft 


^ u 

CO CD 


boy 

0^ 


2 

r 


Sq. In. 


cfl 

5 






Ins. 




Sq.In. 


Sq.In. 


Lbs. 


Lbs. 


Pr.Ct. 


Pr. Ct. 


Lbs. 


Xbs. 


Lbs. 


A 


8Q-38 


100.5 


7.19 


7.60 


*773° 


80280 


25.8 


30.9 


28000000 


38700 


36700 


B 


98.38 


100 


4 


7.19 


7.78 


+5650 


78830 


20.8 


34-1 


2830000c 


40600 


37500 


C 


107.38 


118 


5 


6.73 


7.21 


43900 


77840 


20.4 


42.5 


30000000 


393°° 


36700 


D 


ni-75 


125 





7.27 


7.78 


^756o 


79270 


22.3 


43-2 


28500000 


36100 


337oo 


E 


116.25 


130 





7.38 


7.96 


45820 


79250 






30500000 


39300 


36400 






F 


120.63 


i34 


8 


7.21 


7-55 


49440 


81660 


24.1 


39-0 


28800000 


393oo 


37500 


G 


125-13 


i30 


7 


7.06 


7.67 


3959° 


79690 


24.4 


45-5 


30300000 


38000 


35000 


H 


I34-I3 


149.4 


7.28 


7.78 


3947o 


78650 


21 .0 


28.3 


30800000 


374oo 


35000 



TRANSVERSE STRENGTH OF STEEL. 



The following table gives the results of tests of a number of 
steel I beams, made in the Laboratory of Applied Mechanics 
of the Mass. Institute of Technology. 



TRANSVERSE STRENGTH OF STEEL. 



497 







Mo- 




Break- 


Mo- 
dulus of 


Modulus 
of 




No. of 


Depth. 


ment of 


Span. 


ing 


Rup- 


Elasticity 




Test. 


Inches. 


Inertia. 


Feet and 


Load. 


ture 


per 


Remarks. 






Ins. 


Inches. 


Lbs. 


per 

Sq. In. 
Lbs. 


Sq. In. 
Lbs. 




290 


7 


38.00 


14'' 6" 


10500 


42874 


29030000 


From Phoenix Co. 


293 


8 


57.11 


14' 6" 


14200 


44270 


29410000 


" " " 


2 95 


9 


81.34 


14' 6" 


16700 


40200 


29890000 


( < 11 a 


337 


6 


24.86 


14' 7" 


8200 


44900 


28170000 


N. J. Iron & Steel Co. 


34o 


7 


39 - 6 3 


12' 11" 


12000 


42100 


27480000 


" " tl " 


343 


8 


51.67 


14' 7" 


14900 


46400 


29040000 


c< tt (i a 


631a 


10 


134.00 


14' 0" 


24200 


3 8 5°° 


28400000 


Carnegie Steel Co. 


638 


10 


134.00 


14' 0" 


25100 


395oo 


29300000 


( ( l C CI 


674 


10 


129.00 


14' 0" 


24900 


41300 


27450000 


11 (I c c 


675 


10 


131.20 14' 0" 


25600 


41700 


27850000 


a tt tt 



In Heft IV of the Mitth. der Materialpriifungsanstalt in 
Zurich are given the following results of tests of the transverse 
strength of ten steel plate girders : 



Depth of 


Span. 
(Inches.) 


Modulus of 


Modulus of 


Web. 
(Inches.) 


Rupture. 
(Lbs. per 


Elasticity. 
(Lbs. persq.in.) 






sq. in.) 




19.76 


177.17 


53325 


29193660 


19.76 


177-17 


55316 


27430380 


15-75 


141-73 


55174 


26662500 


15-75 


I4L73 


55316 


28738620 


19.69 


177.17 


53325 


29193560 


19.69 


177.17 


55316 


27430380 


23.62 


2I2.6o 


57591 


28795500 


23.62 


212.60 


52472 


28155600 


27.56 


248 . 03 


54320 


27529920 


27.56 


248 . 03 


53041 


2875284O 



498 APPLIED MECHANICS. 



COLD CRYSTALLIZATION OF IRON AND STEEL. 

The question of cold crystallization of wrought-iron and 
steel is one that has been agitated from the earliest times, and, 
although Kirkaldy tried to dispose of it finally by offering evi- 
dence showing that it does not exist, nevertheless we find the 
same old question cropping out every little while, and although 
the bulk of the evidence is admitted to be against it, and-, as it 
seems to the writer, there is no evidence in its favor, we find 
every now and then some one who thinks that certain observed 
phenomena can be explained in no other way. 

The most usual phenomenon which cold crystallization is 
called upon to explain is the crystalline appearance of the 
fracture of some piece of wrought-iron or steel that has been 
in service for a long time, and which has, as a rule, been sub- 
jected to more or less jars or shocks. The cases most fre- 
quently cited are those of axles of some sort which have been 
broken, and, in the case of which, the fracture has had a crys- 
talline appearance, and where samples cut from the other parts 
of the axle and tested have shown a fibrous fracture. The 
assumption has therefore been made that the iron was origi- 
nally fibrous, and that crystallization has been caused by the 
shocks or the jarring to which it has been subjected in the 
natural service for which it was intended. 

Kirkaldy showed (see his sixty-six conclusions) that when 
fibrous iron was broken suddenly, or when the form of the 
piece was such as not to offer any opportunity for the fibres to 
stretch, the fibres always broke off short, and the fracture was 
at right angles to their length, and hence followed the crystal- 
line appearance ; whereas if the breaking was gradual, and the 
fibres had a chance to stretch, they produced a fibrous appear- 
ance : iri short, he claimed that the difference between the crys- 
talline or the fibrous appearance of the fracture was only a 



COLD CRYSTALLIZATION OF IRON AND STEEL. 499 



difference of appearance, and not a change of internal structure 
from fibrous to crystalline. 

The facts that Kirkaldy showed in this regard are generally 
acknowledged to-day, and doubtless answer by far the greater 
part of those who claimed cold crystallization at the time that 
he wrote, and also a great many of those who claim its exist- 
ence to-day. 

But it is easy, if suitable means be taken, to distinguish 
cases of crystalline appearance of fracture from cases where 
there are actual crystals in the piece ■; and it is rather about 
those cases where the iron near the fracture actually contains 
distinct crystals that what discussion there is to-day that is 
worth considering takes place. 

The number of such cases is, of course, small, but every 
once in a while some one is cited, and the claim is put forward 
that the iron was originally fibrous, and that these crystals 
must therefore have been produced without heating the iron 
to a temperature where chemical change is known to occur. 

Inasmuch as the one who claims the existence of cold crys- 
tallization is announcing a theory which is manifestly opposed 
to the well-known chemical law that crystallization requires 
freedom of molecular motion, and hence, can only take place 
from solution, fusion, or sublimation, it follows that the burden 
of proof rests with him, and before he can substantiate his 
theory in any single case he must prove beyond the possibility 
of doubt, i°, that the iron or steel was originally fibrous, i.e., 
not only that fibrous iron was used in manufacturing the pieces, 
but also that it had not been overheated during its manufac- 
ture, and, 2°, that it has never been overheated during its period 
of service. 

It is because the writer is not aware of any case where these 
two circumstances have been proved to hold that he says that 
he knows of no evidence for cold crystallization. In this con- 
nection it is not worth while to quote very much of the exten- 



500 APPLIED MECHANICS. 



sive literature on the subject ; hence only a little of the most 
modern evidence will be given here. 

On page 1007 et seq. of the report of tests on the govern- 
ment testing-machine at Watertown Arsenal for 1885 is given 
an account of a portion of a series of tests upon wrought-iron 
railway axles, and the following is quoted from that report : — 

" This series of axle tests, begun September, 1883, is carried 
on for the purpose of determining whether a change in struc- 
ture takes place in a metal originally ductile and fibrous to a 
brittle, granular, or crystalline state, resulting from exposure 
to such conditions as are met with in the ordinary service of a 
railway axle. 

" Twelve axles were forged from one lot of double-rolled 
muck-bars, and in their manufacture were practically treated 
alike. Each axle was made from a pile composed of nine bars, 
each 6 in. wide, f in. thick, and 3 ft. 3 in. long, and was finished 
in four heats, two heats for each end. 

" The forging was done by the Boston Forge Company in 
their improved hammer dies, which finish the axle very nearly 
to its final dimensions. 

" Two axles were taken for immediate test, to show the 
quality of the finished metal before it had performed any rail- 
way service, and serve as standards to compare with the 
remaining ten axles, to be tested after they had been in 
use. 

" The axles are in use in the tender-trucks of express loco- 
motives of the Boston and Albany Railroad. Mr.. A. B. Under- 
bill, superintendent of motive-power, contributes the axles and 
furnishes the record of their mileage." 

The results of some measurements of deflection are given 
concerning one of the axles in tender 134, after it had run 
95000 miles ; and then follows : — 

" Regarding the axle for the time being as cylindrical, 3.96 



COLD CRYSTALLIZATION OF IKON AND STEEL. 50 1 

inches diameter, the modulus of elasticity by computation will 
be 28541000 pounds. 

"Applying this modulus to the deflections observed in rear 
axle of the rear trucks of tender No. 150, the maximum fibre 
strain is found to be 9935 pounds per square inch when the 
tender was partially loaded, and 14900 pounds per square inch 
when fully loaded. 

" Taken together, the tensile and compressive stresses, 
which are equal, amount to 19870 and 29800 pounds per square 
inch respectively, as the range of stresses over which the metal 
works. 

" This definition of the limits of stresses must be regarded 
as approximate. There are influences which tend to increase 
the maximum fibre strain, such as unevenness of the track, the 
side thrust of the wheel-flanges against the rails. On the other 
hand, the inertia of the axle, particularly under high rates of 
speed, would exert a restraining influence on the total deflec- 
tion. 

" Nine tensile specimens were taken from each axle ; three 
from each end, including the section of axle between the box 
and wheel bearings, and three from the middle of its length. 
They are marked M.B., with the number of the axle ; also a 
sub-number and letter to indicate from what part of the axle 
each was taken. 

" The tensile test-pieces showed fibrous metal, and generally 
free from granulation. 

" The muck-bar had a higher elastic limit and lower tensile 
strength, and less elongation than the axles. The moduli of 
elasticity of the two are almost identical. 

" Between loads of 15000 and 25000 pounds per square inch 
the muck-bar had a modulus of elasticity of 29400000 pounds, 
the axles (average of all specimens) between 5000 and 20000 
pounds per square inch was 29367000 pounds. Individually 



502 APPLIED MECHANICS. 

the axles showed the modulus of elasticity to be substantially 
the same in each." 

Two specimens were subjected to their maximum load and 
removed from the testing-machine before breaking in order to 
see whether the straining followed by rest will cause any 
change. 

"•It does not appear from these tests that 95000 miles 
run has produced any effect on the quality of the metal." 

On page 1619 et seq. of the Report for 1886 is given an 
account of the tests made on some more of these axles which 
had run 163 138 miles, and the following is quoted from that 
account : — 

" Specimens from muck-bar axle No. 4 after the axle had 
run 1 63 1 38 miles. 

" Comparing these results with earlier tests of this series, the 
tensile strength of the metal in this axle is lower, and the 
modulus of elasticity less than shown by the preceding axles. 

" The variations in strength, elasticity, and ductility are no 
greater, however, than those met in different specimens of new 
iron of nominally the same grade, and while apparently there 
is a deterioration in quality, it needs confirmation of a more 
decisive nature from the remaining axles before attributing 
this result to the influence of the work done in service." 

Another set of tests made at Watertown Arsenal is to be 
found on page 1044 et seq. of the Report for 1885. There were 
tested — 

i°. Two side-rods of a passenger locomotive which had been 
in service about twelve years. 

2°. One side-rod of a passenger engine which had been run 
twenty-eight years and eight months. 

3 . One main-rod which had been run thirty-two years and 
eight months in freight and five years in passenger service. 

In none of these tests were there any evidences of crystal- 
lization, as the metal was in all cases fibrous when fractured. 



COLD CRYSTALLIZATION OF IRON AND STEEL. 5°3 



In the report is said : — 

" There are no data at command telling what the original 
qualities of the metal of these bars were : it is sufficient, how- 
ever, to find toughness and a fibrous appearance in the iron to 
prove that brittleness or crystallization has not resulted from 
long exposure to the stresses and vibrations these bars have 
sustained." 

The only other evidence that will be referred to is the paper 
of Mr. A. F. Hill upon the " Crystallization of Iron and Steel," 
contained in the Proceedings of the Society of Arts of the 
Massachusetts Institute of Technology for 1882-83. In this 
article Mr. Hill covers the ground very fully, and distinctly 
asserts that — 

" The fact is that there is at present not a single well- 
authenticated instance of iron or steel ever having become 
crystallized from use under temperatures below 900 F." 

He claims to have investigated a great many cases where 
cold crystallization has been claimed, and to have found, in 
every case where crystals existed, that at some period, of its 
manufacture or working the metal was overheated. He 
says : — 

" That the crystalline appearance of a fracture is not neces- 
sarily an indication of the presence of genuine crystals is proven 
by the well-known fact that a skilful blacksmith can fracture 
fine fibrous iron or steel in such a manner as to let it appear 
either fibrous and silky, or coarse and crystalline, according to 
his method of breaking the bar. On the other hand, where 
there is genuine crystallization, no skill of manipulation will 
avail to hide that fact in the fracture. The most striking- 
illustrations of this that have come under my notice are the 
fractures of the beam-strap of the Kaaterskill, and of the 
connecting-rod of the chain-cable testing-machine at the Wash- 
ington Navy Yard. The photographs of both fractures are 
submitted to you, and the similarity of their appearance is 



504 APPLIED MECHANICS. 

most singular. Yet what a difference in the development of 
the longitudinal sections by acid treatment, which are also 
presented to you. 

" In the Kaaterskill accident the fractures of both the 
upper and lower arms of the strap were found to be short and 
square. The appearance of the fractured faces showed no 
trace of fibre, and was altogether granular. Yet the longitudi- 
nal section, taken immediately through the break, and devel- 
oped by acid treatment, shows the presence of but few and 
small crystals, and the generally fibrous character of the iron 
used in the strap. 

" In the connecting-rod of the chain-cable testing-machine 
we find the crystalline appearance of the fracture less, if any- 
thing, than that of the beam-strap, while the development of 
the longitudinal section by acid treatment reveals most beauti- 
fully, in this case, the thoroughly crystalline character of the 
metal. As is well known, this rod, after many years of service, 
finally broke under a comparatively light strain, and having all 
along been supposed to have been carefully made, and from 
well-selected scrap, its intensely crystalline structure, as re- 
vealed by the fractures, has done service for quite a number of 
years as piece de resistance in all the ' cold-crystallization ' 
arguments which have been served up in that time." 

He then goes on to say that he cut the rod in a longitudinal 
direction, and treated the section with acid ; that some of the 
crystals shown are so large as to be discernible with the naked 
eye ; that the treated section furnished incontrovertible evi- 
dence that the rod, aside from the fact of being badly dimen- 
sioned anyhow, was made of poor material,. badly heated, and 
msufriciently hammered, all records, suppositions, and asser- 
tions to the contrary notwithstanding ; that there are a large 
number of crystals composed of a substance, presumably a 
ferro-carbide, which is not soluble in nitric acid, and is found 
in steel only ; that the deduction from the large amount of this 



COLD CRYSTALLIZATION OF IRON AND STEEL. 505 

substance is that the pile was formed of rather poorly selected 
scrap, with steel scrap mixed in ; that evidences of bad heating 
are abundant throughout ; and that the strongest evidence 
against the presumption that these crystals were formed during 
the service of the rod, or while the metal was cold, is found in 
the groupings of the crystals during their formation, as shown 
in the tracing developed by the acid ; that they are not of the 
same chemical composition, the lighter parts containing much 
more carbon than the darker ones ; it is therefore pretty evi- 
dent that with the grouping of the crystals a segregation of 
like chemical compounds took place, and this of course would 
have been impossible in the solid state. He then cites an 
experiment he made, in which he took a slab of best selected 
scrap weighing about 200 pounds and forged it down to a 
3-inch by 3-inch square bar, one-half being properly forged, 
and the other half being exposed to a sharp flame bringing it 
quickly to a running heat, keeping it at this heat some time, 
and then hammering lightly and then treating it a second time 
in a similar manner ; the result being, that while no difference 
was discernible in the appearance of the two portions, when 
cut and treated with acid the portion that was properly made 
showed itself to be a fair representative of the best quality of 
iron, while in the other portion the crystallization was strongly 
marked, the majority of the crystals being large and well 
developed. 

He also says : — 

" The fact is, all hammered iron or steel is more or less 
crystalline, the lesser or greater degree of crystallization de- 
pending altogether upon the greater or lesser skill employed 
in working the metal, and also largely upon the size of the 
forging. Crystallization tends to lower very sensibly the elastic 
limit of iron and steel, and therefore hastens the deterioration 
of the metal under strain. It is for this reason that large aid 
heavy forgings ought to be, and measurably are, excluded as 



506 APPLIED MECHANICS. 

much as possible from permanent structures. In machine con- 
struction we cannot do without them, and must therefore 
accept the necessity of replacing more or less frequently the 
parts doing the heaviest work." 

The evidence given above seems to the writer to be suffi- 
cient, and to warrant the conclusions stated on pages 475, 476. 

EFFECT OF TEMPERATURE UPON THE RESISTING PROPER- 
TIES OF IRON AND STEEL. 

Much the best and most systematic work upon this subject 
has been done at the Watertown Arsenal, and an account of it 
is to be found in " Notes on the Construction of Ordnance^ 
No. 50," published by the Ordnance Department at Washing- 
ton, D. C, U.S.A. 

Other references are the following: — 

Sir William Fairbairn: Useful Information for Engineers. 
Committee of Franklin Institute: Franklin Institute Journal. 
Knutt Styffe and Christer P. Sandberg: Iron and Steel. 
Kollman: Engineering, July 30, 1880. 
Massachusetts R. R. Commissioners' Report of 1874. 
Bauschinger: Mittheilungen, Heft 13, year 1886. 

A summary of the Watertown tests, largely quoted from 
the above-mentioned report, will be given here, and then a few 
remarks will suffice for the others. 

The subjects upon which experiments were made at Water 
town were the effect of temperatures upon — 

i°. The coefficient of expansion. 

2°. The modulus of elasticity. 

3 . The tensile strength. 

4 . The elastic limit. 

5 . The stress per square inch of ruptured section* 

6°. The percentage contraction of area. 

7 . The rate of flow under stress. 

8°. The specific gravity. 



EFFECT OF TEMPERATURE ON IRON AND STEEL. 5°7 

9°. The strength when strained hot and subsequently rup- 
tured cold. 

10°. The color after cooling. 
n°. Riveted joints. 



I . THE COEFFICIENTS OF EXPANSION. 

These were determined from direct measurements upon the 
experimental bars, first measuring their lengths on sections 
35 inches long, while the bars were immersed in a cold bath of 
ice-water, and again measuring the same sections after a period 
of immersion in a bath of hot oil. 

The range of temperature employed was about 210 degrees 
Fahr., as shown by mercurial thermometers. 

Observations were repeated, and again after the steel bars 
had been heated and quenched in water and in oil. 

The average values are exhibited in the following : — 

TABLE I. 
First Series of Bars. 





Chemical Composition. 


Coefficients of Expansion 


Metal. 








per Degree Fahr., per 




c. 


Mn. 


Si. 


Unit of Length. 


Wrought-iron. 








. OOOO067302 


Steel. 


.0 9 


.11 






OOOO067561 




.20 




45 






OOOOO66259 




•31 




57 






OOOO065149 




•37 




70 






OOOOO66597 




•5i 




58 


.02 




0000066202 




•57 




93 


.07 




0000063891 




•71 




58 


.08. 




OOOO064716 




.81 




56 


•17 




OOOO062167 




.89 




57 


.19 




OOOO062335 




•97 




80 


.28 




OOOO061700 


Cast- (gun) iron. 










OOOOO59261 


Drawn copper. 








OOOOO91286 



5o8 



APPLIED MECHANICS 



Subsequent determinations of the coefficient of expansion 
of a second series of steel bars gave — 

TABLE II. 



Chemical Composition. 


Coefficients of Expansion 














per Degree Fahr., per 


C. 


Mn. 


Si. 


S. 


P. 


Cu. 


Unit of Length. 


•17 


1 -13 


.023 


.122 


.079 


.04 


.0000067886 


.20 


.69 


.037 


•13 


.078 


.26 


.OOOO068567 


.21 


1.26 


.08 


.14 


•059 


.00 


. OOOOO67623 


.26 


1.07 


.11 


.096 


.08 


•047 


.OOOOO67476 


.26 


1.26 


.07 


.112 


.06 


.038 


.OOOO067102 


.26 


1.28 


.07 


•115 


.062 


.035 


.0000067175 


' .28 


1.23 


.09 


.168 


.09 


.178 


.OOOO067794 


• 43 


•97 


•05 


.08 


.096 


.024 


.0000066124 


•43 


1.08 


.037 


.08 


.114 


•233 


.0000066377 


•53 


•75 


.10 


.078 


.087 


.174 


.OOOO064181 


•55 


1.02 


•05 


.078 


.12 


.15 


.OOOO066122 


•72 


• 70 


.18 


•07 


•13 


•23 


. OOOO064330 


•72 


• 76 


.20 


.056 


.086 


.186 


. OOOO063080 


•79 


.86 


.21 


.084 


•093 


.096 


.OOOO063562 


1.07 


.07 


•13 


.01 


.018 


.006 


.OOOO061528 


1.08 


.12 


.19 


.011 


.02 


trace 


.OOOO061702 


1. 12 


.10 


.09 


.013 


.018 


trace 


.OOOO060716 


1. 14 


.10 


•15 


trace 


.018 


trace 


.0000062589 


1. 17 


.10 


• IO 


trace 


.018 





.0000061332 


I.3I 


•13 


.19 


.011 


.026 


trace 


.OOOO061478 



Ten bars of the first series were now heated a bright cherry- 
red and quenched in oil at 8o° Fahr., the hot bars successively 
raising the temperature of the oil to about 240 Fahr., the bath 
being cooled between each immersion. 

The behavior of the bars under rising temperature, when 
examined for coefficients of expansion, seemed somewhat 
erratic, the highest temperature reached being 235 ; but this 
behavior was subsequently explained by the permanent changes 
in length found when the bars were returned to the cold bath. 



EFFECT OF TEMPERA TURE ON IRON AND STEEL. 509 

Generally the bars were found permanently shortened at the 
close of these observations. 

The bars were again heated bright cherry-red and quenched 
in water at 50 to 55 Fahr., the water being raised by the 
quenching to no° to 125 Fahr. 

After resting 72 hours, measurements were taken in the cold 
bath, followed by a rest of 18 hours, when they were heated 
and measured in the hot bath, after which they were measured 
in the cold bath ; the maximum temperature reached with the 
hot bath being 233°./ Fahr., erratic behavior occurring still. 

They were next heated in an oil bath at 300 Fahr., and 
kept at this temperature 6 hours, then cooled in the bath ; 15 
hours later they were heated to243° Fahr., and again measured 
hot, and then cold. These downward readings showed the 
quenched in water bars to have their coefficients elevated 
above the normal, as shown in the following table, these 
being the same steel bars as in Table I, and in the sarr.e 
order : — 

TABLE III. 



Coefficients of Expansion 
per Degree per Unit of 
Length. 


Apparent Shortening of Bars 
Due to Six Hours at 300 
Fahr., and the following 
Immersion in the Hot Bath. 


.OOOO067641 


— .0006 


. OOOO066622 


.0002 


.OOOO066985 


.OOl6 


.OOOO067377 


.0023 


.OOOO069776 


— .0004 


.OOOOO6704I 


.0082 


. OOOO066939 


.0064 


.OOOOO6879O 


.0054 


.OOOOO72906 


•OO55 


.OOOOO71578 


.0048 



5IO APPLIED MECHANICS. 

Finally the bars were annealed by heating bright red and 
cooling in pine shavings, the effect of which was to approxi- 
mately restore the rate of expansion to the normal, as shown 
by Table I for these ranges of temperature. 

2°. MODULUS OF ELASTICITY. 

These were obtained with the first series of bars at atmos- 
pheric temperatures, and at higher temperatures, up to 495° 
Fahr. 

There occurred invariably a decrease in the modulus of 
elasticity with an increase in temperature, and, in the case of 
the specimens tested, the low carbon steels showed a greater 
reduction in the modulus than the high carbon steels, the 
first specimen having a modulus of elasticity at the minimum 
temperature 30612000, and at the maximum 27419000, while 
the last specimen had at the minimum temperature 29126000, 
and at the maximum 27778000. 

3°. TENSILE STRENGTH. 

The tests were made upon the first series of steel bars, 
wrought-irons marked A and B, a muck-bar railway axle, and 
cast-iron specimens from a slab of gun-iron. 

The specimens were o".798 diameter, and 5" length of stem, 
having threaded ends i".2$ diameter. 

Wrought-iron A was selected because it was found very hot 
short at a welding temperature. It had been strained with a 
tensile stress of 42320 pounds per square inch seven years 
previous to being cut up into specimens for the hot tests. 

The specimens while under test were confined within a 
sheet-iron muffle, through the ends of which passed auxiliary 
bars screwed to the specimens, the auxiliary bars being secured 
to the testing-machine. 



EFFECT OF TEMPERATURE ON IRON AND STEEL. 5 II 

The heating was done by means of gas-burners arranged 
below the specimen and within the muffle. 

The temperature of the test-bar was estimated from the 
expansion of the metal, observed on a specimen length of six 
inches, using the coefficients which were determined at lower 
temperatures, as hereinbefore stated, assuming there was a 
uniform rate of expansion. 

Access to the specimen for the purpose of measuring the 
expansion was had through holes in the top of the muffle. 
The temperature was regulated by varying the number of gas- 
burners in use, the pressure of the gas, and also by means of 
diaphragms placed within the muffle for diffusing the heat. 

The approximate elongations under different stresses were 
determined during the continuance of a test from measurements 
made on the hydraulic holders of the testing-machine, at a 
convenient distance from the hot muffle, correcting these 
measurements from data obtained by simultaneous micrometer 
readings made on the specimen and the hydraulic holders at 
atmospheric temperatures. 

While it does not seem expedient in one series of tests to 
obtain complete results upon the tensile properties at high 
temperatures, yet, incidentally, much additional valuable infor- 
mation may be obtained while giving prominence to one or 
more features. 

From these elongations the elastic limits were established 
where the elongations increased rapidly under equal incre- 
ments of load. Proceeding with the test until the maximum 
stress was reached, recorded as the tensile strength, observing 
the elongation at the time, then, when practicable, noting 
the stress at the time of rupture." 

For the detailed tables of tests the student is referred to 
the " Notes on the Construction of Ordnance." 

The elastic limits and tensile strengths are computed in 
pounds per square inch, both on original sectional areas of the 



512 APPLIED MECHANICS. 

specimens and on the minimum or reduced sections, as meas- 
ured at the close of the hot tests. 

From the results it appears that the tensile strength of the 
steel bars diminishes as the temperature increases from zero 
Fahr., until a minimum is reached between 200 and 300 Fahr., 
the milder steels appearing to reach the place of minimum 
strength at lower temperatures than the higher carbon bars. 

From the temperature of this first minimum strength the 
bars display greater tenacity with increase of temperature, until 
the maximum is reached between the temperatures of about 
400 to 650 Fahr. 

The higher carbon steels reach the temperature of maximum 
strength abruptly, and retain the highest strength over a lim- 
ited range of temperature. The mild steels retain the increased 
tenacity over a wider range of temperature. 

From the temperature of maximum strength the tenacity 
diminishes rapidly with the high carbon bars, somewhat less 
so with mild steels, until the highest temperatures are reached, 
covered by these experiments. 

The greatest loss observed in passing from 70° Fahr. to the 
temperature of first minimum strength was 6.5 per cent at 
295 ° Fahr. 

The greatest gain over the strength of the metal at yo° was 
25.8 per cent at 460 Fahr. 

The several grades of metal approached each other in 
tenacity as the higher temperatures were reached. Thus steels 
differing in tensile strength nearly 90000 pounds per square 
inch at 70 , when heated to 1600 Fahr. appear to differ only 
about 10000 pounds per square inch. 

The rate of speed of testing which may modify somewhat 
the results with ductile material at atmospheric temperatures 
has a very decided influence on the apparent tenacity at high 
temperatures. 

A grade of metal which, at low temperatures, had little 



EFFECT OF TEMPERATURE ON IRON AND STEEL. 5 13 

ductility, displayed the same strength whether rapidly or slowly 
fractured from the temperature of the testing-room up to 6oo° 
Fahr. ; above this temperature the apparent strength of the 
rapidly fractured specimens largely exceeded the others. 

At 1410 Fahr. the slowly fractured bar showed 33240 
pounds per square inch tensile strength, while a bar tested in 
two seconds showed 63000 pounds per square inch. 

Cast-iron appeared to maintain its strength with a tendency 
to increase until 900 Fahr. is reached, beyond which the 
strength diminishes. Under the higher temperatures it devel- 
oped numerous cracks on the surface of the specimens preced- 
ing complete rupture. 

4°. ELASTIC LIMIT. 

The report says of this that it appears to diminish with in- 
crease of temperature. Owing to a period of rapid yielding with- 
out increase of stress, or even under reduced stress, the elastic 
limit is well defined at moderate temperatures with most of the 
steels. 

Mild steel shows this yielding point up to the vicinity of 
500 ; in hard steels, if present, it appears at lower temperatures. 

The gradual change in the rate of elongation at other times 
often leaves the definition of the elastic limit vague and doubt- 
ful, especially so at high temperatures. The exclusion of de- 
terminable sets would in most cases place the elastic limit below 
the values herein given. 

In approaching temperatures at which the tensile proper- 
ties are almost eliminated exact determinations are correspond- 
ingly difficult, the tendency being to appear to reach too high 
values. 

5°. STRESS ON THE RUPTURED SECTION. 

This, generally, follows with and resembles the curve of 
tensile strength. 



514 APPLIED MECHANICS. 

Specimens of large contraction .of area, tested at high 
temperature, have given evidence on the fractured ends of 
having separated at the centre of the bar before the outside 
metal parted. 

Elongation under Stress* 

Although the metal is capable of being worked under the 
hammer at high temperatures, it does not then possess sufficient 
strength within itself to develop much elongation, general 
elongation being greatest at lower temperatures. 

Greater rigidity exists under certain stresses at intermedi- 
ate temperatures than at either higher or lower temperatures. 

Thus one of the specimens tested at 569 Fahr. showed less 
elongation under stresses above 50000 pounds per square inch 
than the bars strained at higher or lower temperatures. 

Two other specimens showed a similar behavior at 315 and 
387 respectively, and likewise other specimens. 

In bars tested at about 200 to 400° Fahr. there are dis- 
played alternate periods of rigidity and relaxation under in- 
creasing stresses, resembling the yielding described as occur- 
ring with some bars immediately after passing the elastic 
limit. 

The repetition of these intervals of rigidity and relaxation 
is suggestive of some remarkable change taking place within 
the metal in this zone of temperature. 

6°. PERCENTAGE CONTRACTION OF AREA. 

This varies with the temperature of the bar ; it is somewhat 
less in mild and medium hard steels at 400° to 6oo° than at 
atmospheric temperatures. 

Above 500 or 6oo° the contraction increases with the 
temperature of the metal ; with three exceptions, which showed 
diminished contraction at 1100 Fahr., until at the highest 
temperatures some of them were drawn down almost to points. 



EFFECT OF TEMPERATURE ON IRON AND STEEL. 5 15 



7 . RATE OF FLOW UNDER STRESS. 

The full effect of a load superior to the elastic limit is not 
immediately felt in the elongation of a ductile metal, and the 
same is true at higher temperatures. 

The flow caused by a stress not largely in excess of the 
elastic limit has a retarding rate of speed, and eventually ceases 
altogether ; whereas under a high stress the rate of flow may 
accelerate, and end in rupture of the metal. 

Hence the apparent tensile strength maybe modified within 
limits by the time employed in producing fracture. 

8°. SPECIFIC GRAVITY. 

In general, the specific gravity is materially diminished in 
the vicinity of the fractured ends of tensile specimens, and this 
diminution takes place in the different grades of steel, in bars 
ruptured under different conditions of temperature, stress, and 
cortti'action of area. 



9 BARS STRAINED HOT, AND SUBSEQUENTLY RUPTURED COLD. 

The effect of straining hot on the subsequent strength cold 
appears to depend upon the magnitude of the straining force 
and the temperature in the first instance. 

There is a zone of temperature in which the effect of hot 
straining elevates the elastic limit above the applied stress, and 
above the primitive value, and if the straining force approaches 
the tensile strength, there is also a material elevation of that 
value when ruptured cold. These effects have been observed 
within the limits of about 335 and 740 Fahr. 

After exposure to higher temperatures there occurs a 
gradual loss in both the elastic limit and tensile strength, and 
generally a noticeable increase in the contraction of area. 



5 l6 APPLIED MECHANICS. 



IO . COLOR AFTER COOLING. 

This was not sensibly changed by temperatures below 200°. 
After 300 the metal was light straw-colored ; after 400 , deep 
straw; from 500 to 6oo°, purple, bronze-colored, and blue; 
after 700 , dark blue and blue black. 

After 8oo° some specimens still remained dark blue. After 
heating above about 8oo° the final color affords less satisfactory 
means of approximately judging of the temperature, the color 
remaining a blue black, and darker when a thick magnetic 
oxide is formed. 

At about 1100 the surface oxide reaches a tangible thick- 
ness, a heavy scale of c/'.ooi to o".oo2 thickness forming as 
higher temperatures are reached. The red oxide appears at 
about 1500 . 

11°. IN THE TESTS OF RIVETED JOINTS 

of steel boiler-plates at temperatures ranging from 70 to about 
700 Fahr. the indications of the tensile tests of plain bars were 
corroborated. 1 

Joints at 200 Fahr. showed less strength than when cold ; 
at 250 and higher temperatures the strength exceeded the 
cold joints ; and when overstrained at 400 and 500 there was 
found, upon completing the test cold, an increase in strength. 

Rivets which sheared cold at 40000 to 41000 pounds per 
square inch, at 300 Fahr. sheared at 46000 pounds per square 
inch ; and at 6oo° Fahr., the highest temperature at which the 
joints failed in this manner, the shearing-strength was 42130 
pounds per square inch. 

In addition to the work at Watertown which has just been 
detailed two other matters will be referred to here. 



EFFECT OF TEMPERATURE ON IRON AND STEEL. 5 17 

1°. It is well known that wrought-iron and steel are very- 
brittle at a straw heat and a pale blue, as shown by the fact that 
when the attempt is made to bend a specimen at these tempera- 
tures it results in cracking it some time before a complete bend- 
ing can be effected, even in the case of metal which is so ductile 
that it can be bent double cold, red hot, or at a flanging heat, 
witnout showing any signs of cracking. 

2°. Bauschinger defines the elastic limit as the load at which 
the stress is no longer proportional to the strain ; whereas he 
calls stretch-limit (Streckgrenze) the load at which the strain 
diagram makes a sudden change in its direction ; i.e., where 
instead of showing a gradually increasing ratio of strain to stress 
it shows a sudden and rapid increase. 

From his experiments (see Heft 13 of the Mittheilungen, 
year 1886) he draws the following conclusions : — 

(a) That the effect of heating and subsequent cooling in 
lowering both the elastic and the stretch limits in mild steel 
begins at about 66o° Fahr. when the cooling is sudden, and at 
about 840 Fahr. when it is slow, and for wrought-iron at about 
750 with either rapid or slow cooling. 

(b) That the operation of heating above those temperatures, 
and of subsequent slow or quick cooling, is that both the elastic 
and the stretch limit are lowered, and the more so the greater 
the heating ; also, that this effect is greater on the elastic than 
on the stretch limit. 

(c) Quick cooling after heating higher than the above-stated 
temperatures lowers the elastic and the stretch limit, especially 
the first, much more than slow cooling, dropping the elastic 
limit almost immediately at a heat of about 930 and certainly 
at a red heat to nothing or nearly nothing in wrought-iron, and 
in both mild and hard steel, while slow cooling cannot bring 
about such a great drop of the elastic limit, even from more 
than a red heat. 



5i8 



APPLIED MECHANICS. 



Effect of Cold-Rolling on Iron and Steel. — It has already 
been stated, p. 410, that it was discovered independently by 
Commander Beardslee and Professor Thurston, that if a load 
were gradually applied to a piece of iron or steel which exceeded 
its elastic limit, and the piece then allowed to rest, the elastic 
limit and the ultimate strength would thus be increased. This 
may be accomplished with soft iron and steel by cold-rolling or 
cold-drawing, but cannot be taken advantage of in hard iron 
or steel. 

Professor Thurston, who has investigated this matter at 
great length, and made a large number of tests on the subject, 
gives the following as the results of cold-rolling : — 



Increase in 


Per Cent. 




25 to 40 
50 to 80 
80 to 125 

300 to 400 
150 to 425 




Elastic limit (tension, torsion, and transverse), 


Elastic resilience (transverse) 



He also says, in regard to the modulus of elasticity, — 

" Collating the results of several hundred tests, the author 
[Professor Thurston] found that the modulus of elasticity rose, 
in cold-rolling, from about 25000000 lbs. per square inch to 
26000000, the tenacity from 52000 lbs. to nearly 70000, the 
elastic limit from 30000 lbs. to nearly 60000 lbs. ; and the ex- 
tension was reduced from 25 to \o\ per cent. 

"Transverse loads gave a reduction of the -modulus of elas- 
ticity to the extent of about 1 000000 lbs. per square inch, an 
increase in the modulus of rupture from 73600 to 133600, and 
reduction of deflection at maximum load of about 25 per cent. 
The resistance of the elastic limit was doubled, and occurred 
at a much greater deflection than with untreated iron." 

On the other hand, the two steel eye-bars referred to on 



FACTOR OF SAFETY. 



519 



p. 472 show a decrease of modulus of elasticity with increasing 
overstrain. 

Whitworth'' s Compressed Steel. — Sir Joseph Whitworth pro- 
duces steel of great strength by applying to the molten metal, 
directly after it leaves the furnace, a pressure of about 14000 
lbs. per square inch; this being sufficient to reduce the length 
of an eight-foot column by one foot. He claims, according to 
D. K. Clark, to be able to obtain with certainty a strength of 
40 English tons with 30 per cent ductility, and mild steel of a 
strength of 30 English tons with ^ or 34 per cent ductility. 

The following tests were made on the Watertown machine, 
upon some specimens of Whitworth steel taken from a section 
of a jacket which was shrunk upon a wr ought-iron tube, and 
removed from shrinkage by the application of high furnace heat : 

TENSILE TESTS. 



Diameter, 


Tensile 
Strength, 


Elastic Limit, 


Contraction 


Inches. 


lbs. per Sq. In. 


lbs. per Sq. In. 


per cent, 


O 564 


IO3960 


55000 


41.9 


O.564 


90040 


48000 


47-2 


O.564 


IO4200 


57000 


24.6 


O.564 


IOOI20 


57000 


44.6 


O.564 


9304O 


53000 


39-2 


O.564 


IO4160 


60000 


24.6 


O.564 


93160 


47000 


39-2 



COMPRESSIVE TESTS. 



Length, 
Inches. 


Diameter, 
Inches. 


Compressive 

Strength, . 

lbs. per Sq. In. 


Elastic Limit, 
lbs. per Sq. In. 


5 
5 

3-94 
3-94 


O.798 
O.798 
O.798 
O.798 


I02IOO 

89OOO 

IOI600 

I OI 600 


6IOOO 
57000 
53O0O 
54000 



§ 227. Factor of Safety. — In order to determine the 

proper dimensions of any loaded piece, it becomes necessary 



520 APPLIED MECHANICS. 

to fix, in some way, upon the greatest allowable stress per 
square inch to which the piece shall be subjected. 

The most common practice has been to make this some 
fraction of the breaking-strength of the material per square 
inch. 

As to how great this factor should be, depends upon — 

i°. The use to which the piece is to be subjected ; 

2°. The liability to variation in the quality of the material ; 

3°. The question whether we are considering, as the load 
upon the piece, the average load, or the greatest load that can 
by any possibility come upon it ; 

4°. The question as to whether the structure is a temporary 
or a permanent one; 

5°. The amount of injury that would be done by breakage 
of the piece ; 
and other considerations. 

The factors most commonly recommended are, 3 for a dead 
or quiescent load, and 6 for a live or moving load. 

A common American and English practice for iron bridges 
is to use a factor of safety of 4 for both dead and moving load. 
In machinery a factor as large as 6 is desirable when there is 
no liability to shocks ; and when there is, a larger factor should 
be used. 

A method sometimes followed for tension and compression 
pieces is, to prescribe that the stretch under the given load 
should not exceed a certain fixed fraction of the length. This 
requires a knowledge of the modulus of elasticity of the mate- 
rial. 

In the case of a piece subjected to a transverse load, it is 
the most common custom to determine its dimensions in accord- 
ance with the principle of providing sufficient strength ; and 
for this purpose a certain fraction (as one-fourth) of the mod- 
ulus of rupture is prescribed as the greatest allowable safe 
stress per square inch at the outside fibre. Thus, for wrought- 
iron from 10000 to 12000 lbs. per square inch is often adopted 



REPEATED STRESSES. 521 

as the greatest allowable stress at the outside fibre, this being 
about one-fourth of the modulus of rupture. 

The other method for dimensioning a beam is, to prescribe 
its stiffness ; i.e., that it shall not deflect under its load more 
than a certain fraction of the span. This fraction is taken as 

T7T0 t0 "8"0"0- 

This latter method depends upon the modulus of elasticity 
of the beam ; and while it is the most advisable method to 
follow, and as a rule would be safer than the other method, 
nevertheless, in the case of very stiff and brittle material it 
might be dangerous ; hence we ought to know also the break- 
ing-weight and the limit of elasticity of the beam we are to use, 
and not allow it to approach either of these. This precaution 
will be especially important to observe in the case of steel 
beams, which are only now being introduced. 

On the other hand, in moving machinery a factor of safety 
of six is usually required when there is no unusual exposure to 
shocks, as in smooth-running shafting, etc. ; and when there 
are irregular shocks liable to come upon the piece, a greater 
factor is used. 

wohler's results. 

§ 228. Repeated Stresses. — The extensive experiments of 
Wohler for the Prussian government, which were subsequently 
carried on by his successor, Spangenberg, were made to deter- 
mine the effect of oft-repeated stresses, and of changes of 
stress, upon wrought-iron and steel. 

In the ordinary American and English practice, it is cus- 
tomary, in determining the dimensions of a piece, as of a bridge 
member, to ascertain the greatest load which the piece can 
ever be called upon to bear, and to fix the size of the piece in 
accordance with this greatest load. 

Wohler called attention to the fact that the load that would 
break a piece depends upon both the greatest and least load 
that it would ever be called upon to bear. Thus, a tension-rod 



522 APPLIED MECHA: 



which is subjected to alternate changes of load extending from 
20000 to 80000 lbs. would require a greater area for safety than 
one which was subjected to loads varying only between the 
limits of 60000 and 80000 lbs. ; and this would require more 
area than one which was subjected to a steady load of 80000 
lbs. 

Wohler expresses this law as follows, in his " Festigk 
versuche mit Eisen und Stahl." 

•* The law discovered by me, whose universal application 
for iron and steel has been proved by these experiments, is as 
follows : The fracture of the material can be effected by 
variations of stress repeated a great number of times, of 
which none reaches the breaking-limit. The differences of 
the stresses which limit the variations of stress determine the 
breaking-strength. The absolute magnitude of the limiting 
stresses is only so far of influence as, with an increasing stress, 
the differences which bring about fracture grow less 

u For cases where the fibre passes from tension to compres- 
sion and vice versa, we consider tensile strength as positive 
and compressive strength as negative ; so that in this case the 
difference of the extreme fibre stresses is equal to the greatest 
tension plus the greatest compression." 

Besides the ordinary tests of tensile, compressive, shearing, 
and torsional strength, he made his experiments mainly on the 
following two cases : — 

1 °. Repeated tensile strength ; the load being applied and 
wholly removed successively, and the number of repetitions 
required for fracture counted. 

2°. Alternate tension and compression of equal amounts 
successively applied, the number of repetitions required for 
fracture being counted. 

In making these two sets of tests, he made the first set in 
two ways : — 

(a) By applying direct tension. 



LAUNHARDTS FORMULA. $2$ 

(b) By applying a transverse load, and determining the 
greatest fibre stress. 

The second set of tests was made by loading at one end a 
piece of shaft fixed in direction at the other, and then causing 
it to revolve rapidly, each fibre passing alternately from tension 
to an equal compression, and vice versa. 

He also tried a few experiments where the lower limit of 
stress was neither zero nor equal to the upper limit, with a 
minus sign, also some experiments on torsion, on shearing, 
and on repeated torsion. 

When Wohler had made his experiments, and published his 
results, there were a number of attempts made by different 
persons to deduce formulae which should depend upon these 
experiments for their constants, and which should serve to deter- 
mine the breaking-strength for any given variation of stresses. 

Only two of these formulae will be given here, viz. : 

i° That of Launhardt for one kind of stress, 

2° That of Weyrauch for alternate tension and compression. 

LAUNHARDT 'S FORMULA. 

The constants used in this formula are : 

i°. t, the carrying-strength (Tragfestigkeit) of the material 
per unit of area, which is the same as the tensile strength as 
determined by the ordinary tensile testing-machine. 

2°. u, the primitive breaking-strength (Ursprungsfestigkeit), 
i.e., the greater t stress per unit of area of which the piece can bear, 
without breaking, an unlimited number of repetitions, the load 
being entirely removed between times. These two quantities 
have been determined experimentally by Wohler; and it is the 
object of Launhardt's formula to deduce, in terms of t, u, and the 
ratio between the greatest and least loads to which the piece is 
ever subjected, the value a of the breaking-strength per unit of 
area when these loads are applied. 



524 APPLIED MECHANICS. 




Let the greatest stress per unit area be a. 
the least stress per unit area be c. 

Plot the values of - as abscissae, and those of a as ordinates, 
making OA = u (since when - = o, a = u), OC=i, and CB = t 

(since when -=i, a = t). Then will any curve 

which passes through the points A and B have 
for its ordinates values of a that will satisfy the 
conditions that when c = o, a = u, and when c = t, 
a = L By assuming for this curve, the straight 

line AB we obtain DE = AO+FE = AO + (BG)q£ , and hence 

a = u + (t-u)-, (i) 

which is Launhardt's formula. 

Moreover, if we denote by max L the greatest load on the en- 
tire piece, and by min L the least, we shall have 

c min L 
a max V 
Hence 

x min L , N 

this being in such a form as can be used. Or we may write it 
thus: 



)/— wmin L ) . 



this being the more common form. 

The values of the constants as determined by Wohler's experi- 
ments, and the resulting form of the formula for Phoenix axle-iron 
and for Krupp cast-steel, have already been given in § 172. 



WEYRAUCH'S FORMULA. 525 

In the same paragraph are given the corresponding values of 
b, the safe working-strength, the factor of safety being three. 

weyrauch's formula for alternate tension and 
compression. 

The constants used in this formula are : 

i°. u, the primitive breaking-strength, which has been already 
defined. 

2 . s, the vibration breaking-strength (Schwingungsfestigkeit) 
i.e., the greatest stress per unit of area, of which the piece can 
bear, without breaking, an unlimited number of applications, 
when subjected alternately to a tensile, and to a compressive 
stress of the same magnitude. 

He lets a = greatest stress per unit of area, c= greatest stress 
of the opposite kind per unit of area. If a is tension, c is com- 
pression, and vice versa. 

Plot the values of - as abscissae, and those of a as ordinates, 

a ' 

making OA=u (since when — = 1, a=w), OC = i, and CB = s 
(since when — = 1 , a = s) . Then will any curve 

Or 

which passes through the points A and B 
have for its ordinates values of a that will 
satisfy the conditions that when c=o, a=u, 
and when c = s, a=s. 

By assuming for this curve the straight line AB we obtain 

DE=DF-FE=DF-{BG) ( ^ i and hence 

a=u-(u-s)- f (4) 

which is the Weyrauch formula. 




526 APPLIED MECHANICS. 

Moreover, if we write 

c max U 
a max L ' 

where max L= greatest load on the piece, and max L' = great est 
load of opposite kind, so that, if L is tension, U shall be com- 
pression, and vice versa, we shall have 

max U 
a=W -( W - 5) ma^T' <*> 

this being in a form suitable to use, the more common form being 

\ u—sm&xL'} 

a=u]i- =-L (6) 

[ u max L \ v 

The values of the constants as determined from Wohler's 
experiments, and the resulting form of the formulae for Phoenix 
axle-iron and for Krupp cast-steel, are given in § 176. 

GENERAL REMARKS. 

In each case the value of a given by the formula (3) or (6) 
is the breaking-strength per unit of area. 

If either of these values of a be divided by 3, we have, accord- 
ing to Weyrauch, the safe working-strength. 

WOHLER'S EXPERIMENTAL RESULTS. 

Wohler himself made his tests upon the extremes of fibre 
stresses of which a piece could bear, without breaking, an 
unlimited number of applications. He gives, as a summary of 
these results, the following : — 

In iron, — 

Between -f 16000 lbs. per sq. in. and —16000 lbs. per sq. in. 
« 4.30000 " " " o " " 

+ 44000 " " " +24000 " " 

In axle-steel, — 

Between +28000 lbs. per sq. in. and —28000 lbs. per sq. in. 
" +48000 " " " o " " 

+ 80000 " " " +35000 " " 



WO'BLER'S EXPERIMENTAL RESULTS. 527 

In untempered spring steel, — 

Between +50000 lbs. per sq. in. and o lbs. per sq. in. 

+ 70000 " " " +25000 " " 

+ 80000 " " " +40000 " " 

« +90000 " " " +60000 " " 

For shearing in axle-steel, — 

Between +22000 lbs. per sq. in. and —22000 lbs. per sq. in. 
« +38000 " " " o " " 

This table would justify the use, in Launhardt's and Wey- 
rauch's formulae, of the following values of u and s ; viz., — 
In iron, — 

u = 30000 lbs. per sq. in., 
s = 16000 lbs. per sq. in. 



In axle steel. 



u = 48000 lbs. per sq. in., 
s = 28000 lbs. per sq. in. 



In untempered spring steel, — 

u = 50000 lbs. per sq. in. 

And it would require, that if, with these values of u f and the 
values of t given in §§ 172 and 176, we put 

c = 24000 

in Launhardt's formula for iron, we ought to obtain approxi- 
mately 

a = 44000 ; 

and if we put c = 35000 in that for steel, we should obtain 
approximately 

a = 80000. 



5^8 APPLIED MECHANICS. 



FACTOR OF SAFETY. 

We have seen that Weyrauch recommends, to use with 
Wohler's results, a factor of safety of three for ordinary bridge 
work and similar constructions. 

Wohler himself, however, in his " Festigkeits versuche mit 
Eisen und Stahl," says, — 

i°. That we must guard against any danger of putting on 
the piece a load greater than it is calculated to resist, by assum- 
ing as its greatest stress the actually greatest load that can 
ever come upon the piece ; and 

2°. This being done, that the only thing to be provided for 
is the lack of homogeneity in the material. 

3°. That any material which requires a factor of safety 
greater than two is unfit for use. This advice would hardly be 
accepted by engineers, however. 

He also claims that the reason why it is safe to load car- 
springs so much above their limit of elasticity, and so near 
their breaking-load, is, that the variation of stress to which they 
are subjected is very inconsiderable compared with the greatest 
stress to which they are subjected. 

GENERAL REMARKS. 

It is to be observed, — 

i°. The tests were all made on a good quality of iron and 
of steel, consequently on materials that have a good degree of 
homogeneity. 

2°. The specimens were all small, and the' repetitions of load 
succeeded each other very rapidly, no time being given for the 
material to rest between them. 

3°. No observations were made on the behavior of the piece 
during the experiment before fracture. 



SHEARING-STRENGTH OF IRON AND STEEL. 529 

4 . As long as we are dealing only with tension, we can say 
without error that 

c _ min L m 
a max L 

but as soon as both stresses or either become compression, if 
the piece is long compared with its diameter, we cannot assert 
with accuracy the above relation, nor that 

c _ maxZ' , 
a maxZ 

and hence results based on these assumptions must be to a 
certain extent erroneous. 

5°. When a piece is subjected to alternate tension and com- 
pression, it must be calculated so as to bear either : thus, if 
sufficient area is given it to enable it to bear the tension, it may 
not be able to bear the compression unless the metal is so dis- 
tributed as to enable it to withstand the bending that results 
from its action as a column. 

While Worrier's tests were mostly confined to ascertaining 
breaking-strengths, the later experimenters upon this subject, 
especially Prof. Bauschinger at Munich, Mr. Howard at the 
Watertown Arsenal, and Prof. Sondericker at the Mass. Institute 
of Technology, have all undertaken to study the elastic changes 
developed in the material by repeated stresses, and also, to some 
extent, the effect upon resistance to repeated stress, of flaws, cf 
indentations, and of sudden changes of section, including sharp 
corners. 

They all agree in the conclusion that flaws and indentations 
(even though very slight) and sharp corners, including keyways, 
reduce the resistance to repeated stress very considerably. 

A brief account will be given of some of their principal con- 
clusions. 



53° APPLIED MECHANICS. 



bauschinger's tests on repeated stresses. 

Bauschinger's tests upon repeated stress include work upon 
the properties of metals at or near the elastic limit. Of the 
properties which he enumerates, the following will be quoted 
here : 

(a) The sets within the elastic limit are very small, and in- 
crease proportionally to the load, while above that point they 
increase much more rapidly. 

(b) With repeated loading, inside of the elastic limit,- dropping 
to zero between times, we find each time the same total 
elongations. 

(c) While within the elastic limit the elongations remain 
constant as long as the load is constant ; with a load above the 
elastic limit the final elongations under that load are only reached 
after a considerable length of time. 

(d) If by subjecting a rod to changing stresses between an 
upper and lower limit, of which at least the upper is above the 
original elastic limit, the latter were either unchanged or lowered, 
or if, in the case of its being raised, it were to remain below the 
upper limit, then the repetition of such stresses must finally end 
in rupture, for each new application of the stress increases the 
strain; but if both limits of the changing stress are and 
remain below the elastic limit, the repetition will not cause 
breakage. 

(e) Bauschinger says that by overstraining, the stretch limit 
is always raised up to the load with which the stretching was 
done; but in the time of rest following the unloading the stretch 
limit rises farther, so that it becomes greater than the max- 
imum load with which the piece was stretched, and this rising 
continues for days, months, and years; but, on the other hand, 
that the elastic limit is lowered by the overstraining, often to 
zero; and that a subsequent rest gradually raises it until it 
reaches, after several days, the load applied, and in time 



EXPERIMENTS WITH A REPEATED TENSION MACHINE. 53 1 



rises above this ; that, as a rule, the modulus of elasticity 
is also lowered under the same circumstances, and is also 
restored by rest, and rises after several years above its 
original magnitude. 

(f) By a tensile load above the elastic limit the elastic 
limit for compression is lowered, and vice versa for a compres- 
sive load ; and a comparatively small excess over the elastic 
limit for one kind of load may lower that for the opposite- 
kind down to zero at once. Moreover, an elastic limit which 
has been lowered in this way is not materially restored by a 
period of rest — at any rate, of three or four days. 

(g) With gradually increasing stresses, changing from 
tension to compression, and vice versa, the first lowering of 
the elastic limit occurs when the stresses exceed the original 
elastic limit. 

(h) If the elastic limit for tension or compression has been 
lowered by an excessive load of the opposite kind, i.e., one ex- 
ceeding the original elastic limit, then, by gradually increasing 
stresses, changing between tension and compression, it can 
again be raised, but only up to a limit which lies considerably 
below the original elastic limit. 

EXPERIMENTS WITH A REPEATED TENSION MACHINE. 

Bauschinger states that in 1881 he acquired a machine 
similar to that used by Wohler for repeated application of a 
tensile stress. 

The plan of the experiments which he made with it, and 
which are detailed in the 13th Heft of the Mittheilungen, is as 
follows : 

From a large piece of the material there were cut at least 
four, and sometimes more, test-pieces for the Wohler machine. 
One of them was tested in the Werder machine to determine 
its limit of elasticity and its tensile strength ; the others were 



53 2 APPLIED MECHANICS. 

tested in the Wohler machine, so arranged that the upper limit 
of the repeated stress should be, for the first specimen, near 
the elastic limit ; for the second, somewhat higher, etc., the 
lower limit being in all cases zero. 

From time to time the test-pieces, after they had been sub- 
jected to some hundred thousands, or some millions, of repeti- 
tions, were taken from the Wohler machine and had their limits 
of elasticity determined in the Werder machine. 

The tables of the tests are to be found in the Mittheilungen, 
and from them Bauschinger draws the following conclusions : 

i°. With repeated tensile stresses, whose lower limit was 
zero, and whose upper limit was near the original elastic limit, 
breakage did not occur with from 5 to 16 millions of repeti- 
tions. 

Bauschinger says that in applying this law to practical cases 
we must bear in mind two things : (a) that it does not apply 
when there are flaws, as several specimens which contained flaws, 
many of them so small as to be hardly discoverable, broke with 
a much smaller number of repetitions ; (b) another caution is 
that we should make sure that we know what is really the origi- 
nal elastic limit, as this varies very much with the previous 
treatment of the piece, especially the treatment it received 
during its manufacture, and it may be very small, or it may be 
very near the breaking-strength. 

2°. With oft-repeated stresses, varying between zero and an 
upper stress, which is in the neighborhood of or above the 
original elastic limit, the latter is raised even above, often far 
above, the upper limit of stresses, and the higher the greater 
the number of repetitions, without, however,, its being able to 
exceed a known limiting value. 

3 . Repeated stresses between zero and an upper limit, 
which is below the limiting value of stress which it is possible 
for the elastic limit to reach, do not cause rupture ; but if the 
upper limit lies above this limiting value, breakage must occur 
after a limited number of repetitions. 



EXPERIMENTS WITH A REPEATED TENSION-MACHINE. 533 

4°. The tensile strength is not diminished with a million 
repetitions, but rather increased, when the test-piece after hav- 
ing been subjected to repeated stresses is broken with a steady 
load. 

5°. He discusses here the probability of the time of forma- 
tion of what he considers to be a change in the structure of 
the metal at the place of the fracture. 

Besides the above will be given the numerical values which 
Bauschinger obtained for carrying strength and for primitive 
safe strength as average values. 

i°. For wrought-iron plates : 

t = 49500 lbs. per sq. in. 
u = 28450 " " " " 

2°. For mild-steel plates (Bessemer) : 

/ = 62010 lbs. per sq. in. 

u= 34140 " " " " 

# t 

3 . For bar wrought-iron, 80 mm. by 10 mm. : 

t = 57600 lbs. per sq. in. 
u= 31290 " " " " 

4 . For bar wrought-iron, 40 mm. by 10 mm. ; 

/ = 57180 lbs. per sq. in. 
u = 34140 " " " " 

5°. For Thomas-steel axle : 

t = 87050 lbs. per sq. in. 
u = 42670 " " « " 



6°. For Thomas-steel rails : 



/ = 8/^-90 lbs. per sq. in. 
u = 39820 " " " " 



534 APPLIED MECHANICS. 

7°. For Thomas-steel boiler-plate : 

£=57600 lbs. per sq. in. 
u = 34140 " " " " • 

For Thomas -steel axle, and Thomas-steel rails, Bauschinger's 
obtained for the vibration breaking-strength the same values as 
those for primitive breaking-strength. His experiments on the 
other five materials, however, give lower values for s than for 
u. These values will not be quoted here, however, because they 
were obtained from experiments upon rotating bars of rectangular 
section transversely loaded. 

EXPERIMENTS UPON ROTATING SHAFTING SUBJECTED TO TRANS- 
VERSE LOADS, BY PROF. SONDERICKER. 

Accounts of these tests are to be found in the Technology 
Quarterly of April, 1892, and of March, 1899. In every case the 
(transverse) loads were so applied, that a certain portion, greater 
than ten inches in length, was subjected to a uniform bending- 
moment. At various times, the shaft was stopped, the load was 
removed, then replaced, and again removed, and measurements 
made of the strains and sets. The diameter of the shaft was, 
in every case, approximately one inch. Some extracts from the 
paper of March, 1899, will be given. The investigations were 
conducted along two lines. 

i°. The determination of elastic changes, resulting from 
the repeated stresses, and the influence of such changes in pro- 
ducing fracture. 

2 . The influence of form, flaws, and local conditions generally 
in causing fracture. 

Accurate measurements of the elastic strains, and sets were 
made at intervals during each test. Characteristic curves of set 
indicate the general character of the 
changes which occurred in the set, the 
abscissas being the number of revo- 
lutions, and the ordinates the amount 
of the set, a is the characteristic curve 




EXPERIMENTS WITH A REPEA TED TENSION MA CHINE. 535 



for wrought-iron, and also occurred in one land of soft steel. No 
change is produced until the elastic limit is reached, and then 
the change consists in a decrease of set. b is the characteristic 
curve for all the steels tested with the single exception mentioned. 
It is the reverse of the preceding, beginning commonly below the 
elastic limit, and consisting of an increase of set; rapid at first, 
but finally ceasing. Under heavy loads, the increase of set is 
very rapid, and ceases comparatively quickly. Accompanying 
the change of set there is a change in the elastic strain in the 
same direction but much smaller in amount. From the fact that 
these changes finally cease, we conclude that, if of sufficiently 
small magnitude, they do not necessarily result in fracture. 

The table on page 536 gives a number of his results. 

Regarding these results he says : 

i°. In several cases, changes would have been detected under 
smaller stresses had observations been taken. 

2 . Changes of set may be expected to begin at stresses vary- 
ing from J to J of the tensile strength. 

3 . The set does not appear to have a notable influence in 
causing fracture until it reaches o".ooi or o".oo2 in a length of 
ten inches. 

4°. The effect of rest is to decrease the amount of set. In 
most cases, however, the set lost is soon regained, when the bar is 
again subjected to repeated stress, especially in the case of the 
harder steels. 

Prof. Sondericker also cites a few experiments to determine 
the loss of strength due to indentations, grooves, and keyways. 
In one case, the result of cutting a groove around the steel shaft 
about 0^.003 deep was a loss of strength of about 40 per cent, 
while similar results were obtained with indentations, and with 
square shoulders. He also cites the case of two pieces of steel 
shafting united by a coupling, where the result of cutting the 
necessary keyways in the shafts caused, apparently, a loss of 
about 50 per cent. 



536 



APPLIED MECHANICS. 





' 


Tensile Prop- 
erties of the 
Metal. 




Revolu- 
tions at 




Maxi- 








Stress 


which 


Revolu- 
tions. 


mum 




i 


Material. 


Elastic 
Limit 


Tensile 
St'gth 


per 

Sq. In. 


Change 

was First 

Ob- 


Observed 
Sets. 


Remarks* 






per 

Sq. In. 
Lbs. 


per 

Sq. In. 

Lbs. 




served. 












Lbs. 






Inches. 




D 


Wt.-Iron 


15700 


45080 


30000 


42300 


86400 


< .01200 


Broke at one end at 
shoulder, and at 
other where arm 
was attached. 


40 




24000 


50700 


24000 
26000 


1500000 


1500000 
2427000 




Broke. 


1 


' ' 


25900 


51390 


26000 


2214000 


2285000 


.00026 


Broke near center. 


2 


' * 


25900 51390 


32000 


486000 


486000 


.00136 


Broke at mark burn- 


















ed by electric cur- 


















rent. 


3 




23400 


50510 


24000 
24000 
25000 
26000 
27000 
28000 


6593000 


6593000 
4059000 
8962000 
3932000 
8155000 
589000 


.00037 
.00011 
.00016 
.00022 
.00037 
.00038 


Broke at shoulder. 


4 


" 


23400 


50510 


28000 


2506000 


2506000 


.00042 


Broke at center. 


33 


Steel 


24800 47400 


32000 


85900 


89750 


.00771 


Broke outside of arm 
near bearing; color 
blue black. 


34 


' ' 


24800! 47400 


32000 


103500 


1 1 6600 


.00832 


Do. 


21 




30400 62590 


32000 


4395000 


4395000 


.00029 












34000 




8339000 


.00032 












36000 




4627000 


.00041 












36000 




1428000 


.00008 


After resting unload- 
ed 18 days. 








38000 




3769000 


.00023 




| 








40000 




4523000 


.00054 












42000 




505000 


.00072 


Broke near shoulder. 


54 




42000 


63130 


45000 


163000 


163000 


.00312 


Broke at shoulder. 


50 




23200 


7376o 


30000 
35000 


339000 


339000 
16400 


.00100 
.00282 


Not broken. 


25 




38300 


78010 


40000 
42000 


5031000 


5031000 
2483000 


.00028 
.00046 


Broke at shoulder. 


26 




38300 


fSoio 


40000 
42000 


20838000 


20838000 
3311000 


.00037 
.00044 


Not broken. 


18 




50000 


81010 


36000 


6463000 


6982000 


.00052 


Broke where arm was 

attached. 
Broke at shoulder. 


19 




50000 


81010 


36000 


7252000 


7686000 


.00069 


20 




50000 


81010 


34000 
36000 
38000 
38000 

40000 


21221000 


21221000 

13577000 

2263000 

9237000 

932000 


.00028 
.00067 
.00113 
.00116 

.00177 


After resting 6 mos. 

unloaded. 
Broke at shoulder. 


53 


" 


58000 


96580 


50000 


24000 


146500 


.00249 


Broke at shoulder. 


58 




58000 


96580 


45000 
50000 


50100 


50100 
156900 


.00020 
.00289 


Broke near middle. 


29 




54000 


104480 


40000 
42000 
44000 
46000 
48000 
50000 
50000 


5257000 


5257000 
7125000 
4626000 
6760000 
4965000 
170000 
1000 


.00046 
.00067 
.00100 
.00145 
.00196 
.00197 
.00203 


After 24 days rest un- 
loaded ; not broken. 


55 




50000 


104830 


35000 
40000 
50000 


276900 


276900 

237900 

22530 


.00060 
.00274 
.00615 


Broke near shoulder; 
color dark straw. 


57 


" 


50000 


104830 


60000 


14300 


14900 


.00768 


Broke near shoulder; 




1 " 












color dark blue. 



EXPERIMENTS WITH A REPEATED TENSION MACHINE. 537 

TESTS OF ROTATING SHAFTING UNDER TRANSVERSE LOAD, BY 
MR. HOWARD AT THE WATERTOWN ARSENAL. 

A large number of tests of this character have been made at 
the Watertown x\rsenal. A few extracts will be given from the 
remarks of Mr. Howard upon the subject, which may be found 
in the Technology Quarterly of March, 1899, as follows: 

"In the Watertown tests, two principal objects have been 
in view, namely, to ascertain the total number of repetitions of 
stresses necessary to cause rupture, and to observe through what 
phases the physical properties of the metal pass prior to the 
limit of ultimate endurance. The Watertown tests have included 
cast-iron, wrought-iron, hot and cold rolled metal, and steels 
ranging in carbon from 0.1 per cent to 1.1 per cent, also milled 
steels. The fibre-stresses have ranged from 10000 pounds 
per square inch on the cast-iron bars up to 60000 pounds per 
square inch on the higher tensile-strength steel bars. 

The speed of rotation was from 400 per minute up to 2200 
per minute, in different experiments. Observations were made 
on the deflection of the shafts, and on the sets developed. It 
was early observed that intervals of rest were followed by tem- 
porary reduction in the magnitude of the sets. In the Report 
of Tests of Metals of 1888, he says the deflections tend to 
diminish under high speeds of rotation, when the loads exceed 
the elastic limit of the metal, and tend to cause permanent sets; 
but, on the other hand, when the elastic limit is not passed, the 
deflections are the same within the range of speeds yet experi- 
mented upon. 

Efforts were inaugurated at this time to ascertain the effect 
of repeated alternate stresses on the tensile properties of the 
metal, and it appeared that such treatment tended to raise the 
tensile strength of the metal before rupture ensued. 

Concerning the limit of indefinite endurance to repeated 
stress we know but very little. In most experiments rupture 
occurs after a few thousand repetitions, so high have been the 



53 3 



APPLIED MECHANICS. 



applied stresses. Examples are not uncommon in railway prac- 
tice of axles having made 200000000. rotations. In order to 
establish a practical limit of endurance, indefinite endurance, 
if we choose to call it so, our experimental stresses will need to 
be somewhat lowered, or new grades of metal found. 

The following table which accompanied the Watertown 
Arsenal Exhibit at the Louisiana Purchase Exposition gives a 
summary of some of the repeated stress tests upon three different 
grades of steel: 

STEEL BARS. 
Tensile Tests and Repeated Stress Tests on Different Carbon Steels. 







Tensile Tests. 


Repeated Stress Tests. 










Mechan- 


Maxi- 










Elastic 


Ten- 


Elon- 


Con- 


ical 


mum 




Mechan- 




Description. 


Limit 


sile 


gation 


trac- 


Work at 


Fiber 


Number 


ical Work 






per 


St'gth 


in 4 


tion of 


Rupture 


Stress 


of Rota- 


at Rup- 






Sq. In. 


per 


Ins. 


Area 


per 


per 


tions at 


ture per 
Cu. In. 






Lbs. 


Sq. In. 


Per ct. 


Per ct. 


Cu.In. 


Sq. In. 


Rupture. 








Lbs. 






Ft.-lbs. 


Lbs. 




Ft.-lbs. 
















( 60000 
j 50000 


6470 


32835 
















17790 


62635 





1 7 Carbon steel . 


51000 


68000 


33-5 


5i.9 


982 


\ 45000 

i 4OOOO 

1 35000 

I 30000 

60000 

50000 


70400 

293500 

5757920 

♦23600000 

12490 

93160 


201960 

665290 

9992390 

♦29500000 

63387 

328000 





55 Carbon steel. 


57000 


106100 


16.2 


18.7 


1.047 


. 45000 
40000 
35000 
. 30000 
60000 
55000 


166240 

45535o 

900720 

♦19870000 

37250 

9379o 


476900 

1032130 

1563125 

♦24838000 

189044 
■ 399780 





82 Carbon steel. 


63000 


142250 


8.5 


6.5 


888 


50000 
j 45000 
1 4OOOO 


213150 
605460 


750465 
1736910 
















♦17560000 


♦40973000 
















I 35000 


♦19220000 


♦33635000 



* Not ruptured. 
GENERAL REMARKS. 

That the amount of detailed information regarding repeated 
stresses is small compared with what is needed will be evident 
when we cpnsider the number of cases in which metal is subjected 
to such stresses in practice, among which are shafting, connecting- 
rods, parallel rods, propeller-shafts, crank-shafts, railway axles, 
rails, riveted and other bridge members, etc. In the case of 



TORSIONAL STRENGTH OF WROUGHT IRON AND STE^L. 539 



some of them, notably, railway axles, attempts have been made 
to base specifications for the material upon such tests as have 
become available upon repeated stresses. 

§ 229. Shearing-strength of Iron and Steel. — Some of 
the most common cases where the shearing resistance of iron 
and steel is brought into play are: 

i°. In the case of a torsional stress, as in shaftmg. 

2 . In the case of pins, as in bridge-pins, crank- pins, etc. 

3 . In the case of riveted joints. 

The so-called apparent outside fibre-stress at fracture, as 
determined from experiments on torsional strength, is found 
to be not far from the tensile strength of the metal, and is, of 
course, greater than the shearing-strength, for the same reasons as 
render the modulus of rupture greater than the actual outside 
fibre-stress at fracture in transverse tests. 

Moreover, the shearing strength of wrought-iron rivets is 
shown by experiment to be about f the tensile strength of the 
rivet metal. 

In regard to cast-iron, Bindon Stoney found the shearing and 
tensile strength about equal. 

The cases where shearing comes in play in wrought-iron and 
steel will therefore be treated separately. 

§ 230. Torsional Strength of Wrought-iron and Steel. — 
The method formerly followed, and in use by some at the present 
day, was to compute the strength of a shaft from the twisting- 
moment only, neglecting the bending, but varying the working- 
strength per square inch to be used according to the character 
of the service. It is generally the fact, however, that when 
shafting is running the pulls of the belts create a bending back- 
wards and forwards, bringing the same fibre alternately into 
tension and compression; and this is combined with the shearing- 
stresses developed due to the twisting- moment alone. At the 
two extremes of these general cases are : 

i°. The case when the portion of a shaft between two hangers 



540 APPLIED MECHANICS. 

has no pulleys upon it, and when the pulls on the neighboring 
spans are not so great as to deflect this span appreciably. That 
is a case of pure torsion: and if the shaft. is running smoothly, 
with no jars or shocks, and no liability to have a greater load 
thrown upon it temporarily, we may compute it by the usual 
torsion formula, given in § 212; using for breaking-strength of 
wrought-iron and steel the so-called apparent outside fibre-stress 
at fracture as determined from torsional tests, and a factor of 
safety six, and such a proceeding will probably give us a reasonable 
degree of safety. 

2 . The case when, pulleys being placed otherwise than near 
the hangers, the belt-pulls are so great that the torsion becomes 
insignificant compared with the bending, and then it would be 
proper to compute our shaft so as not to deflect more than y^-g- 
of its span under the load, or better, not more than ygV : °f 
course we should compute also the breaking transverse load, and 
see that we have a good margin of safety. 

In other cases, the methods pursued, the first two of which 
are incorrect, have been 

i°. By using the ordinary torsion formula combined with a 
large factor of safety. 

2 . By computing the shaft also for deflection, and providing 
that its deflection shall not exceed xtqq or Tr J-g- of its span. 

This, however, neglects the torsion, and also the rapid change 
of stress upon each fibre from tension to compression. 

3 . By using the formula of Grashof or of Rankine for com- 
bined bending and twisting, with the constants that have been 
derived from experiments on simple tension or simple torsion. 

The results given on pages 544 and 545 are from pieces of 
shafting of considerable length. As has been stated, the so-called 
"apparent outside fibre-stress at fracture " ^appears to be not very 
far from the tensile strength of the material, and the torsional 
modulus of elasticity appears to be from three- eighths to two- 
fifths of the tensile modulus of elasticity. 



TORSIONAL STRENGTH OF WROUGHT-IRON AND STEEL. 54 1 

Under certain circumstances the bending may have the 
greatest influence, while the twisting may be predominant in 
others, or their influence may be equally divided. Which of these 
is the case will depend upon the location of the hangers and of 
the pulleys, the width of the belts, etc., etc. 

As to the formulas which take into account both twisting and 
bending, there are two, both of which are based upon the theory 
of elasticity. The first, which is the most correct from a theo- 
retical point of view, is that given by Grashof and other writers 
on the theory of elasticity, and is 

' I { 2M 2m J 

where M\ = greatest bending-moment ; 
M 2 = greatest twisting-moment; 
r = external radius of shaft; 

/= moment of inertia of Section about a diameter; 
/ = greatest "allowable stress at outside fibre; 
m = a constant depending on the nature of the material. 
In the case of iron or steel the value of m is often taken as 4, 
though it is, in most cases, nearer 3. When m = 4 we have 



/ 



j\ %Mi + WM! 2 +M 2 2 



The other formula, which is also based upon the theory of 
elasticity, but which is not as correct, is that given by Rankine, 

and is 



/ 



~\ m 1 Wm 1 2 +m 2 2 \. 



With a view to determine the behavior of shafting under a 
combination of twisting and bending, suitable machinery was 
erected in the engineering laboratories of the Mass. Institute of 
Technology, and a number of tests were made. 



542 



APPLIED MECHANICS. 



The principal points of the method of procedure are the 
following, viz. : 

ist. The shaft under test is in motion, and is actually driving 
an amount of power which is weighed on a Prony brake. 

2d. A tr nsverse load is applied which may be varied at the 
option of the experimenter, and which is weighed on a platform 
scale. 

3d. The proportion between the torsional and transverse 
loads may be adjusted to correspond with the proportion be- 
tween the power transmitted and the belt-pull sustained by a 
shaft in actual use. 

4th. Tests are made not only of breaking-strength, but also 
angle of twist and deflection under moderate loads are measured. 

The following table will give the results of the tests on 
iron shafts, and they will then be discussed : 



No. 

of 
Test. 


Time 
of 

run- 
ning, 

min- 
utes. 


Total 
revolu- 
tions. 


H. P. 

trans- 
mitted. 


"JMTi, 

max. 
bending 
moment. 
In.-lbs. 


max. 
twisting 
moment. 
In.-lbs. 


A, 

max. 
bend. 

fibre 
stress. 


/a, 

max. 
twist. 

fibre 
stress. 


Grashof. 


Ran- 
kine. 


Diam. 
ins. 


8 

9 

10 
11 
12 
13 

15 
16 

17 

19 
20 
21 
22 
23 


37-5 
200 
162 

553 
408 
98 
423 
565 
443 
95i 


7040 

38839 
31641 
108002 
80694 

19333 

82741 
108739 

88208 
185233 


11. 717 
8. 181 

5-291 
4-33i 
6.276 
6.342 
6.283 
6.192 
6.338 
6.283 

14.834 

7.562 
9.972 
15-159 
2-955 


11514.1 

10507.8 
9891.0 
9241.7 
9241.7 
8917. 1 
8917. 1 
8592.5 
8267.8 

3781.5 

8218 

7976 
8917 
8917 
7652 


3926.4 
2656.8 
1714.6 
1399.2 
2027.6 
2028.2 
2029.7 
2031.6 
2026.8 
2029 . 7 

4744 
2394 
3232 
4848 
945 


60024 
54777 
51562 
48179 
48179 
46485 
46485 

44793 
43100 

38503 

84185 
82112 
90793 
90793 
77913 


10234 
6925 
4469 
3647 
5287 
5287 
5290 
5295 
5283 

10333 

24152' 
12188 
16454 
24681 
4811 


62162 
55876 
52062 

48539 
48911 

47245 
47246 
45582 
43914 
41768 

91928 
84819 
94434 
08612 
78314 


61755 
55671 
5^76 
48769 
48769 
47105 
47106 
45436 
43713 
41117 

90368 
83031 
937i6 
97103 
78239 


l".2 5 
l".2 5 
l".2 5 
l".2 5 
l".25 
l".2 5 
l".2 5 
l".25 
l".25 

1" 
1" 
1" 
1" 















TORSIONAL STRENGTH OF WROUGHT-IRON, ETC. 543 

In 19 to 23 inclusive the number of revolutions was small and 
the outside fibre stress at fracture was correspondingly large. 

Two specimens of the \" '.25 shafting and two of the 1" 
were tested for tension, the results being as follows : 

Breaking-strength, per sq. in. 

„ ,. ( No. i • 46800 

i".2 5 diameter j Nq 2 ^ 

Average .... 48333 

1" diameter \ No * * 5868 7 

1 aiameter | No 2 6l8l2 

Average .... 60250 

As to conclusions : 

1st. It is plain from these results that a shaft whose size is 
determined by means of the results of a quick test would be 
too weak, and that our constants should be obtained from tests 
which last for a considerable length of time. 

2d. A perusal of the tables will show that the results ob- 
tained apply more to the bending than to the twisting of a 
shaft, as the transverse load used in these tests was so large 
compared with the twist as to exert the controlling influence. 
This will be plain by a comparison of the values of/",, f i9 
and f. 

3d. Nevertheless, the bending-moments actually used were 
generally less than such as might easily be realized in practice 
with the twisting-moments used. 

4th. It seems fair to conclude that, in the greater part of 
cases where shafting is used to transmit power, as in line-shaft- 
ing or in most cases of head shafting, the breaking is even more 
liable to occur from bending back and forth than from twist- 
ing, and hence that in no such case ought we to omit to 
make a computation for the bending of the shaft as well as the 
twist. 

5th. As to the precise value of the greatest allowable out- 
side fibre stress to be used in the Grashof formula, it is plain 



544 



APPLIED MECHANICS, 



that it is not correct to use a value as great as the tensile 
strength of the iron, and while the tests show that this figure 
should not for common iron exceed 40000 lbs. per square inch, 
it is probable that tests where a longer time is allowed for 
fracture will show a smaller result yet. 

TORSIONAL TESTS OF WROUGHT-IRON. 



I 




Norway Iron. 


! 


Burden's Best. 


V 


C CA 


. c -r 
tn a 




."2 •- 


3 C 


en <n 
OJ3 

U (J 


c x 

V V 
4»J3 


tie 
ex: 
•- o 
S c 


£a 


."2 -5 

t/3 . . 


3 .s 

3 • • 


u 

V 
V 

a 


« . 
O en 

c a 
.Z20 


H . 

a|2 


3'c 

Ho , 

^ C u 

.^ 3 
*" ti r! 

I** 


£w 0. 

v V ■ 
u u tfl 

§£d 


£°d 


w 

s§ 

a 

J Si 


*2 

_j c 
eyi-< 

V . 
v> 

c a 

to (5 


H . 

Sc 

3 a 


3'C 

HO 

V" 4> 5 

IBS 


IS? 

0) w . 
v. v- en 
«i.n.o 

££d 


O.tJ « 

■cw.2 
jyod 

C/3 


s 


Q 


S 


fe 


<5 


Cfl 


Q 


5 


s 


& 


<: 


2.00 


70.40 


72360 


16.50 


46065 


I I 406000 


2.01 


63.8 


85050 


9-5° 


53300 


I 1300000 


2.02 


72.00 


74970 


16.00 


46600 


13215000 


2.01 


59.0 


86400 


8.6a 


54200 


1 1500000 


2.03 


71-30 


72000 


14.00 


43757 


12902000 


2.01 


53-o 


84510 


6.87 


53000 


11200000 


2.02 


70.40 


74520 


17.00 


46321 


12247000 


2.00 


58.8 


87480 


8.40 


55700 


1 1600000 


2.02 


69.80 


72000 


14.25 


45837 


12738000 


2.00 


65.5 


85410 


8.5* 


54400 


11600000 


2.03 


70.30 


74880 


15.50 


4569a 


11361000 


2.01 


60.2 


85590 


8.82 


53700 


I 1200000 


2.03 


7^-13 


74880 


20.00 


44658 


I 1957000 


2.02 


58.5 


85140 


8.05 


52600 


1 1 300000 


2.03 


70.20 


79560 


t6.oo 


48437 


11554000 


2.00 


57-o 


82650 


7-3 1 


52600 


11500000 


2.03 


84 


74880 


i5-5 


45590 


1 1900000 


2.02 


57-8 


86580 


8.54 


53500 


1 1 200000 


1.54 


54 


35100 


12.0 


48950 


9840000 


2.02 


59-5 


86040 


8.61 


53200 


II2000OO 


152 


49 


34200 


11.25 


49600 


11410000 


2.02 


60.0 


87840 


!' 9 l 


54300 


1 130OOOO 


I -53 


S3 


33840 


8.50 


48120 


I I 600000 


2.02 


60.0 


88200 


8.48 


54400 


1 1200000 


i-53 


49 


33840 


ii. 10 


48110 




2.01 


53-3 


87480 


7.85 


54900 


11300000 


J-53 


49 


34920 


14.56 


49650 


I I 840000 


2.01 


59-5 


83970 


8.01 


52700 


I 1400000 


1-52 


S3 


34200 


8.98 


49600 


12480000 


2.01 


59-5 


84780 


8.32 


53200 


1 1 200000 


2.25 


70 


1 1 i960 


5.80 


50060 


11830000 


2.03 


61.0 


83520 


8.98 


50900 


IIIOOOOO 


2.27 


75 


106920 


12.30 


46600 


10900000 


2.00 


63.0 


84050 


9.24 


53500 


II700000 


2.25 


70 


108360 


10.90 


48500 


1 1 800000 


2.02 


60.3 


85950 


7-94 


53100 


11200000 


2.25 


76 


109800 


9.90 


49100 


I 1 700000 


2.01 


60.0 


84600 


8.62 


53100 


I 1400000 


2.23 


7° 


ii 3670 


11.00 


52200 


12000000 


2.02 


61.0 


83520 


8.50 


51600 


IIOOOOOO 


2.26 


70 


107640 


10.00 


47500 


I 1400000 


2.00 


60.5 


86040 


8.80 


54000 


11600000 






Refin 


id Iror 


I. 




2.01 
2.01 


51.0 
36.8 


85680 
87480 


7-35 
8.66 


53700 
54900 


IIOOOOOO 
I 1500000 


2.03 


72.10 


84240 


4.0c 


5"85 


12471000 


2.01 


47.8 


85860 


7.24 


53900 


II300000 


2.03 


71.20 


66960 


1-25 


40646 


12576000 


2.01 


59-4 


85050 


8.92 


53300 


I I 500000 


2.03 


70.50 


70200 


3-5° 


41743 


1 1372000 


2.01 


60.9 


86400 


8-75 


54200 


11300000 


2.03 


72. 10 


72000 


2.30 


43834 


10960000 


2.00 


59-5 


86400 


9.08 


55000 


I 1700000 


2.03 


71.80 


61920 


2.80 


36820 


1 1393000 


2.01 


58.5 


85650 


8.30 


53700 


I I 500000 


2.03 


71-3° 


68760 


2.50 


40887 


1 1360000 


2.00 


58.5 


84870 


7-87 


54200 


I 1800000 


2.03 


69.30 


78120 


2.80 


46453 


12871000 


2.01 


58.3 


87300 


9-45 


54800 


I 1500000 


1.30 


7* -75 


22320 


14.60 


52127 


1 1436000 


2.01 


58.4 


86490 


8-73 


54200 


I 1900000 


1.50 


71.25 


36360 


10.30 


54867 


1 1482000 


2.01 


58.4 


87120 


8.09 


54600 


IIOOOOOO 


1-33 


71-75 


45360 


6.70 


50852 


12359000 


2.00 


59-8 


84870 


8.80 


54000 


I I 700000 


1.50 


79-5o 


32760 


14.60 


49435 


1 107 1000 














2.27 


62 


123800 


4.71 


53920 


12720000 














2.26 


63 


122950 


5.60 


54250 


12230000 














2.29 


64 


124600 


5-24 


52840 


12510000 














2.25 


60 


1 1 6640 


3- 70 


52150 


12190000 














2.25 


61 


121590 


4.40 


54370 


12510000 














2.27 


66 


123750 


5.00 


53890 


12840000 














1.77 


71 


38970 


2. 10 


35790 


1 1 200000 














i-75 


61.1 


56150 


7-i 


53400 


1 1200000 














J -75 


64.0 


5535o 


8.7 


52600 


1 1200000 














1.72 


64-5 


45090 


5-4 


44900 


1 1800000 














»-75 


61.0 


53360 


6.1 


50700 


11500000 














/4 


63.0 


557io 


1.49 


52900 


I T 3OOOOO 















The above tables show the results of tests made in the engi- 
neering laboratories of the Massachusetts Institute of Technol- 



TORSIONAL STRENGTH OF WROUGHT-IRON AND STEEL. 545 

ogy upon the torsional strength of various kinds of wrought-iron. 

The figures in the column headed " Apparent outside fibre-stress " 

Mr 
are obtained from the formula / = -y-, where M = maximum 

twisting-moment, r= outside radius of shaft, and I — polar moment 
of inertia of section. Of course it is not the outside fibre-stress. 



TORSIONAL TESTS OF BESSEMER STEEL. 



to 

u 

C 
»— i 

u 

V 

a 

8 

5 


p. 

'C 


c 

E 

1 
m 

s 


to 
<u 
.g 

c 

to 

»c 

■H 

bo 
G 
O 
h5 

*0 

9 

3 


1 

i 

I 

.2 

1 


ci v 
ft 

•ftd 

"dto H 
O 


& 

bo 
c 

If. 


Apparent Outside Fibre- 
stress Calculated from 
Maximum Twisting- 
Moment, Lbs. per Sq. In. 


O <u 

ft 

10 


>. 

fa* 

mm 

CO 


lO 1 ^ 

i-s 

Ill 


i-75 
i-39 
i-39 

2 .OO 
2 .OO 
2 .02 

1-85 
2 OI 
2 .02 

1-5° 
1.52 

i-53 
1.49 
1.50 
1.50 

i-5° 
1.50 

i-5° 
i-5° 
1.50 
1.50 


60.00 
59-75 
59-5° 
56.00 
55 -oo 
56.00 
44- 00 
55 00 
190.00 
96.00 
94.00 
93.00 
94.00 
60.00 
58.00 

57- 5P 
57.60 
59.00 
58.80 
59 -io 
58.20 
58.00 








66960 
39600 

37620 

101520 

100260 

I I i960 

81240 

I 12 590 
I34IOO 
560OO 
532 80 
532 8o 
5 2 56o 
4392 
44640 
44820 
45816 
45450 
44460 
45000 
4492O 
45540 


63632 

75031 
71250 
64630 
63830 
70630 
65350 
70610 
82860 

84500 
80400 

77300 
74700 
66280 


12418000 

I 12 43OOO 
12594OOO 
I 182OOOO 
IO32OOOO 
1 199OOOO 
IO25OOOO 
I34IOOOO 
I 183OOOO 
12200000 
12200000 
IO70OOOO 
IO9OOOOO 
1 180OOOO 
I I 700000 
I I9OOOOO 
I I9OOOOO 
I I 60OOOO 
I I 700000 
I I 5OOOOO 
1 1 700000 
I I 70OOOO 


11.88 
15.40 
15.00 

7-8 5 

7.87 

10.94 

3-8 7 

5 76 

6.77 

16.30 

i-57o 

16. 10 

15.80 

8.60 

13.30 

11.50 

9.90 

10.80 

10.50 

13.20 

10.80 

11 .40 














3° 
30 
36 
24 
36 
144 

75 
75 
75 
75 
40 
40 
40 
40 
40 
40 
40 
40 
40 


























20160 
19800 
21600 
2 1600 


3040O 
2990O 

31300 
30800 


14400 
17000 
17000 
16000 
18000 
16200 
18000 
18000 


21700 
25600 
25600 
24100 
27200 
24.400 
27200 
27200 


67400 
67600 
69100 
68600 
67100 
67900 
67800 
68700 



546 APPLIED MECHANICS. 

§ 232. Riveted Joints. — The most common way of uniting 
plates of wrought-iron or steel is by means of rivets. It is, 
therefore, a matter of importance to know the strength of such 
joints, and also the proportions which will render their efficien- 
cies greatest j i.e., that will bring their strength .as near as 
possible to the strength of the solid plate. 

In § 177 was explained the mode of proportioning riveted 
joints usually taught, based upon the principle of making all 
the resistances to giving way equal, and assuming, as the modes 
of giving way, those there enumerated. This theory does not, 
however, represent the facts of the case, as — 

i°. The stresses which resist the giving- way are of a more 
complex nature than those there assumed, so that the efficiency 
of a joint constructed in the way described above may not be 
as great as that of one differently constructed ; 

2 . The effects of punching, drilling, and riveting, come in 
to modify further the action ; and 

3 . The purposes for which the joint is to be used, often fix 
some of the dimensions within narrow limits beforehand. 

In order to know, therefore, the efficiency of any one kind 
of joint, we must have recourse to experiment. And here again 
we must not expect to draw correct conclusions from experi- 
ments made upon narrow strips of plate riveted together with 
one or two rivets ; but we need experiments upon joints in wide 
plates containing a sufficiently long line of rivets to bring into 
play all the forces that we have in the actual joint. The greater 
part of the experiments thus far made have been made upon 
narrow strips, with but few rivets. The number of tests of the 
other class is not large, and of those that have been made, the 
greater part merely furnish us information as to the behavior 
of the particular form of joint tested, and do not teach us how 
to proportion the best or strongest joint in any given plates, as 
no complete and systematic series of tests has thus far been 
carried out, though such a series has been begun on the govern- 
ment testing-machine at the Watertown Arsenal. 



RIVETED JOINTS. 54f 



The only tests to which it seems to the writer worth while 
to make reference here are : 

i°. A portion of those made by a committee of the British 
Institution of Mechanical Engineers, inasmuch as, although a 
very large part were made upon narrow strips with but few 
rivets, nevertheless a portion were made upon wide strips. a 

2°. The tests on riveted joints that have been made on the' 
government testing-machine at Watertown Arsenal. 

i°. The account of this series is to be found at intervals 
from 1880 to 1885 inclusive, with one supplementary set in 1888, 
in the proceedings of the British Institution of Mechanical 
Engineers; but as all except the supplementary set has also 
been published in London Engineering, these latter references 
will be given here as follows : 

Engineering for 1880, vol. 29, pages no, 128, 148, 254, 300, 350. 

" 1881, vol. 31, " 427, 436, 458, 508, 588. 

" 1885, vol. 39, " 524. 

" 1885, vol. 40, " 19,43- 
Also, Proc. Brit. Inst. Mechl. Engrs., Oct. 1888. 

2 . The second series, referred to above, or those made on 
the government testing-machine at Watertown Arsenal, are to 
be found in their reports of the following years, viz., 1882, 
1883, 1885, 1886, 1887, and 1895. 

3 . Report of tests of structural material made at the 
Watertown Arsenal, Mass., June, 1891. 

While it is from tests upon long joints that we can derive 
correct and reliable information to use in practice, and hence 
while the experiments already made give us a considerable 
amount of information, nevertheless as the tests have not yet 
been carried far enough to furnish all the information we need, 
and to settle cases that we are liable to be called upon to 
decide, therefore, before quoting the above experiments, a few 
of the rules and proportions more or less used at the present 



54^ APPLIED MECHANICS. 

time, and the modes of determining them, will be first ex- 
plained. 

In this regard we must observe that practical considerations 
render it necessary to make the proportions different when the 
joint is in the shell of a steam-boiler, from the case when it is 
in a girder or other part of a structure. 

In the case of boiler-work, the joint must be steam-tight, and 
hence the pitch of the rivets must be small enough to render 
it so : whereas in girder-work this requirement does not exist ; 
and hence the pitch can, as far as this requirement goes, be 
made greater. 

It is probable, that, with good workmanship, we might be able 
to secure a steam-tight joint with considerably greater pitches 
than those commonly used in boiler-work ; and now and then 
some boiler-maker is bold enough to attempt it. 

Some years ago punching was the most common practice , 
but now drilling has displaced punching to such an extent that 
all the better class of boiler-work is now drilled, and drilling is 
also used to a very considerable extent in girder-work. When 
drilling is used, the plates, etc., to be united should be clamped 
together and the holes drilled through them all together. In 
this regard it should be said : 

i°. When the holes are drilled, and hence no injury is done 
to the metal between the rivet-holes, this portion of the plate 
comes to have the properties of a grooved specimen, and hence 
has a greater tensile strength per square inch than a straight 
specimen of the same plate, as the metal around the holes has 
not a chance to stretch. This excess tenacity may amount 
to as much as 25 per cent in some cases, though it is usually 
nearer 10 or 12 per cent, depending not only on the nature of 
the material, but also on the proportions. 

2°. When the holes are punched, we have, again, a grooved 
specimen, but the punching injures the metal around the hole, 
and this injury is greater the less the ductility of the metal : 
thus, much less injury is done by the punch to soft-steel plates 



EIVETED JOINTS. 549 



than to wrought-iron ones, and less to thin than to thick plates. 
This injury may reach as much as 35 per cent, or it may be 
very small. Besides this, in punching there is liability of crack- 
ing the plate, and of not having the holes in the two plates that 
are to be united come exactly opposite each other. A number of 
tests on the tenacity of punched and drilled plates of wrought- 
iron, and of mild steel, made on the government testing-machine 
at Watertown Arsenal, are given on page 564 et seq. 

The hardening of the metal by punching also decreases the 
ductility of the piece. 

The injury done by punching may be almost entirely re- 
moved in either of the following ways : — 

1 °. By annealing the plate. 

2 . By reaming out the injured portion of the metal around 
the hole ; i.e., by punching the hole a little smaller than is de- 
sired, and then reaming it out to the required size. 

There is a certain friction developed by the contraction of 
the rivets in cooling, tending to resist the giving way of the 
joint ; and some have advocated the determination of the safe 
load upon a riveted joint on the basis of the friction developed, 
instead of on the basis of strength — notably M. Dupuy in the 
Annales des Ponts et Chausees for January, 1895 ; but this 
seems to the author an erroneous and unsafe method of pro- 
ceeding : i°, because tests show that slipping occurs at all 
loads, beginning at loads much smaller than the safe loads on 
the joint ; 2°, because all friction disappears before the break- 
ing load is reached. 

Hence it is safer to disregard friction in designing a tensile 
riveted joint. 

The shearing-strength of the rivets would appear to be 
about two thirds the tensile strength of the rivet metal. 

Before proceeding to give an account of Kennedy's tests, 
and of those made at the Watertown Arsenal, which form the 
principal basis for determining the constants, i.e., the tearing- 
strength of the plate, the shearing-strength of the rivet iron, 



550 APPLIED MECHANICS. 

and the ultimate compression on the bearing surface, it will be 
best to outline the proper method of designing a riveted joint,, 
and for this purpose a discussion of a few cases of tensile riveted 
joints, as given by Prof. Peter Schwamb, will be given by way 
of illustration. 

The letters used will be as follows, viz. : 

d = diameter of driven rivet in inches ; 

/ = thickness of plate in inches ; 

t 1 = thickness of one cover-plate in inches ; 

f = shearing-strength of rivet per square inch ; 

f t = tearing-strength of plate per square inch ; 

f = crushing-strength of rivet or plate per square inch ; 

/ = pitch of rivets in inches ; 

p d = diagonal pitch in inches ; 
/ = lap in inches. 

In every case of a tension-joint we begin by selecting a 
repeating section and noting all the ways in which it may fail. 
It would seem natural, then, to determine the diameter of the 
rivet to be used by equating the resistance to shearing and 
the resistance to crushing, and in some cases it is desirable to 
adopt the resulting diameter of rivet ; but there are also many 
cases where there is good reason for adopting either a larger 
or a smaller rivet, and others where there is good reason for 
determining the trial diameter in some other way. 

Thus we may find that the rivet which presents equal re- 
sistance to shearing and crushing may be too large to be suc- 
cessfully worked, or it may require a pitch too large for the 
purposes for which the joint is to be used; or, on the other 
hand, it may be so small that it would lead to a pitch too 
small to be practicable ; or it might, in a complicated joint, 
where there are a good many ways of possible failing, lead to 
a low efficiency. In all cases, a commercial diameter must be 
selected. 



SINGLE-RIVETED LAP-JOINT. 55 1 

Singie-riveted Lap-joint. — Repeating section containing 
one rivet may fail by — 

i°, shearing one rivet. Resistance = f s % 

4 

2°, tearing the plate. Resistance =f t (p — d)t. 

3°, compression. Resistance = f c td. 

Equating i° and 3 gives d ■=— -j (1) 

A larger rivet will crush, a smaller one will shear. 
The diameter given by (1) will frequently be found to be 
larger than can be successfully worked. 

Equating 2° and 3 gives p — d ( 1 -f- -y], (2) 

* ft' 

Equating 1 ° and 2° gives / = d f 1 -j- — ■£]. (3) 

If the value of d given in (1) is used, then (2) and (3) give 
the same result. If, however, a different value of d is used, 
then the pitch should be determined by (2) for a larger and 
by (3) for a smaller rivet. 

It may be well to note that whenever compression fixes 
the pitch, the computed efficiency 

4-d f c 



P f+fc 

is independent of the diameter of the rivet, and that this is 
the maximum efficiency obtainable with this style of joint. 

SINGLE-RIVETED DOUBLE-SHEAR BUTT-JOINT. 

The combined thickness of the two cover-plates should 
always be greater than t, and, this being the case, we proceed 
as follows : 



55 2 APPLIED MECHANICS. 

Repeating section containing one rivet may fail by — 

I °, shearing one rivet in two places. Resistance = f s —^— m 

2°, tearing the plate. Resistance = f t (p — d)t. 

3°, compression Resistance =f e td. 

2/ f 
Equating i° and 3 gives d= — J J (4) 

A larger rivet will crush, a smaller one will shear. 

The diameter given by (4) is just one half that given by (1), 
r nd will frequently be found to lead to a pitch too small to use in 
practice. In such cases we should use a larger rivet. 

Equating 2 and 3 gives p = d 1 1 + j) . (5) 

Equating i° and 2 gives p = d(i+ — 'j\. (6) 

If the value of d given by (4) be used, then (5) and (6) give 
the same result. If, however, a different value of d be used, then 
the pitch should be determined by (5) for a larger and by (6) for 
a smaller rivet. 

For the diagonal pitch, in the case of staggered riveting, we 
should have, at least, according to Kennedy's sixth conclusion 
(see page 566) 2(p d —d)=$(p—d) and hence p d =%p + $d. 

DOUBLE-RIVETED LAP-JOINTS. 

Repeating section containing two rivets may fail by — 

nd 2 
i°, shearing two rivet sections. Resistance = /, — . 

2 , tearing plate straight across. Resistance =}t(p—d)t. 
3 , compression on two rivets. Resistance —} c (2td). 

Equating i° and 3 gives d=— /. (7) 

7Z js 



EXAMPLE OF A SPECIAL JOINT. 553 

A larger rivet will crush, a smaller one will shear. 

The diameter given by (7) would usually be found too large. 

Equating 2 and 3 gives p = d \i +-r) • (8) 

h 

Equating i° and 2 gives p = d \i H f). (9) 

The pitch should be determined by (8) for a larger and by (9) 
for a smaller rivet than that given by (7). 

For pd we should have, as in the last case, according to 
Kennedy, p d = %p + id. 

EXAMPLE OF A SPECIAL JOINT. 

The joint shown in the cut is one where a part of the rivets 
are in single and a part in double shear. 

Repeating section containing five rivet 
sections may fail by — 

i°, tearing on ab. 

Resistance = f t {p —d)t. 

2 , shearing five rivet sections. 

Resistance = } 8 . 

4 

3 , tearing on ce, and shearing one rivet on ab, 

izd 2 
Resistance =}t(p — 2d)t + /« — . 

4 

4 , tearing on ce, and crushing one rivet. 

Resistance = } t (p — 2d) -f f c tid. 

5 , crushing two rivets and shearing one. 

izd 2 
Resistance = } c (2td) +j s — . 
4 

6°, crushing on three rivets. Resistance = f c (2td+tid). 
7°, crushing three rivets, where h ^ /. 

Resistance =3/^. 



554 APPLIED MECHANICS. 

In this case, we should so proportion the joint that its effi- 
ciency may be determined from its resistance to tearing along ab. 
Hence all its other resistances should be equal to or greater than 
this. 

Hence equate i° and 3 , and calculate the resulting diameter 
of rivet, which will generally be too small, and hence we select 
a larger rivet, so that 3 may be greater than i°. 

Having fixed the diameter of rivet, determine the pitch in 

each of three ways, viz., by equating i° and 2 , by equating 1 0, 

and 6°, and by equating i° and 5 , and adopt the least value of p* 

nd 2 
In this joint as used fad > f 8 — , and hence 6° is greater than 



5°- 



LAP. 



To compute the lap, the following method is a good one. 
Consider the plate in front of the rivet as a rectangular beam 
fixed at the ends and loaded at the middle, whose span=d, 
breadth = t (for cover-plate / x ), depth = A=/— d/2. Assume for 
modulus of rupture j t and for center load W, where 

nd 2 

i°. When rivet fails by single shear W = } s . 

nd 2 
2 . When rivet fails by double shear W = h — . 
J 2 

3 . When rivet fails by crushing and lap in plate is sought 
W=}Jd. 

4°. When rivet fails by crushing and lap in cover-plate is 
sought W=}chd. 



JOINTS IN THE WEB OF A PLATE GIRDER. 



555 



JOINTS IN THE WEB OF A PLATE GIRDER. 



While no experiments on the strength of such joints have 
been published, the constants necessary for use in the ordinary 
method of calculating them are : i°, the allowable outside 
fibre-stress ; 2°, the allowable shearing-stress on the outer 
rivet ; and, 3 , the allowable compression on the bearing- 
surface. 

As an example of the usual method of calculation of such 
a joint, let us consider a chain-riveted butt-joint with two 
covering strips (as shown in the cut) as being a joint in the 
web of a plate girder which has equal 
flanges, and let us determine the allow- 
able amount of bending-moment which 
the web alone (without the flanges) can 
resist. The modifications necessary 
when the flanges are unequal, and 
hence when the neutral axis is not at 
the middle of the depth, will readily 
suggest themselves. 

The stress on any one rivet is pro- 
portional to its distance from the 

neutral axis of the girder, and hence, in this case, from the 
middle of the depth. 

Use the following letters, viz.: 

f t = allowable stress per sq. in. at outer edge of web-plate ; 
f s = allowable shearing-stress per sq. in. on outer rivet ; f c = 
allowable bearing-pressure per sq. in. on outer rivet ; t — thick- 
ness of plate ; h = total depth of web-plate ; h x = total depth 

TTd* 



O O 1 o o / 

o o ! o o I 

o o ! o o / 

o j o o 

o o j o o 

o O o o 

o o j o o 

0000 
\ 



of girder ; d = diameter of driven rivet ; a = 



— area of 



driven rivet section ; r = number of vertical rows on each side ; 
2n = number of rivets in each vertical row ; y x = distance from 



556 APPLIED MECHANICS. 

neutral axis to centre of nearest rivet ; j\ = distance from 
neutral axis to centre of second rivet, etc., etc.; y n =j distance 
from neutral axis to centre of outer rivet. 

Then, for allowable bending-moment, we must take the 
least of the three following, viz : 

i°, that determined from the shearing f s ; 

2°, that determined from the compression f c \ 

3°, that determined from maximum fibre-stress f ti observ- 
ing that if / = greatest allowable fibre-stress in girder, then 

/.=/!■ ; 

To determine these proceed as follows : 

1°. Greatest allowable shear on each outer rivet is 



2fA=f.-j", 



hence allowable stress on rivet at distance y m from neutral axis 



2f s a f s 7td 9 

y n ym ~ 2y n y ™ ' 



and the moment of this stress is 



2U . = fc£ . 



Hence greatest allowable moment on joint for shearing is 



fntfr 

z ^- r \y>'+j>; +?; + ■■■+?»'}■ (i) 

yn 



JOINTS IN THE WEB OF A PLATE GIRDER. 557 



2°. Greatest allowable compression on outer rivet \s fjd \ 
hence allowable stress on rivet at distance^ from neutral axis is 



fdd 



and the moment of this stress is 



y n Jm 



Hence greatest allowable moment on joint for compression is 
2f c tdr 



y* 



I^ 2 +7, 2 +^ + ...+^ 8 (. (2) 



3 . The section of the plate is a rectangle, width t and 
height h, with the spaces where the rivet-holes are cut left out. 
It will be near enough to take for the stress to be deducted on 
account of the rivet-hole at distance y m from neutral axis 



f I y™ \td 



and for its moment 



2ffd , 



Hence greatest allowable moment on joint for tearing is 



558 APPLIED MECHANICS, 






This mode of calculation for (3) would seem to be war- 
ranted from the fact that the rivets do not fill the holes, 
although many deduct only the effect of the holes on the ten- 
sion side, and consider that those on the compression side do not 
weaken the metal. The greatest allowable bending-moment on 
the joint is the smallest of (1), (2), and (3), and it is plain that, in 
order to make the calculation, we need to know what to use 

h 

for/,/, and/, or, since /=/-^-, what to use for //, and 

/; and while /should be determined from the tests on the 
transverse strength of the metal, whether wrought-iron or steel, 
the best evidence we have as to the proper values of/ and/ 
is furnished by the tests on tension-joints, which have already 
been discussed. 

Moreover, we might determine the diameter of rivet by 
equating (1) and (2), but we should generally find it desirable 
to use a larger rivet, and then we should determine the pitch 
by equating (2) and (3) if a larger, or (1) and (2) if a smaller, 
rivet is used. 

Moreover, the rivets in common use in such cases are either 
f " or |" in diameter. 



TESTS OF THE COMMITTEE OF THE BRITISH INSTITUTION OF 
MECHANICAL ENGINEERS. 

The Committee on Riveted Joints of the British Institu- 
tion of Mechanical Engineers consisted of Messrs. W. Boyd, 
W. O. Hall, A. B. W. Kennedy, R. N. J. Knight, W. Parker, 
R. H. Twedell, and W. C. Unwin. 



RIVE TED JOIN TS. 559 



Before beginning operations Prof. Unwin was asked to 
prepare a preliminary report, giving a summary of what had 
already been done by way of experiment, and also to make 
recommendations as to the Course to be pursued in the tests. 

This preliminary report is contained in vol. xxix. of Engineer- 
ing, on the pages already cited. In regard to its recommenda- 
tions it is unnecessary to speak here, as the records of the tests 
show what was done ; but in regard to the summary of what 
had been done, it may be well to say that he gives a list of 
forty references to tests that had been made before 1880, be- 
ginning with those of Fairbairn in 1850, and ending with some 
made by Greig and Eyth in 1879, together with a brief account 
of a number of them. 

Almost all of this work was done, however, with small strips 
with but few rivets, and will not be mentioned here. Inas- 
much, however, as Fairbairn's proportional numbers have been 
very extensively published, and are constantly referred to by 
the books and by engineers, it may be well to quote a portion 
of what Unwin says in that regard, as follows : 

" The earliest published experiments on riveted joints, and 
probably the first experiments on the strength of riveting ever 
made, are contained in the memoir by Sir Wm. Fairbairn in the 
Transactions of the Royal Society. 

" The author first determined the tenacity of the iron, and 
found, for the kinds of iron experimented upon, a mean tenacity 
of 22.5 tons per square inch with the stress applied in the 
direction of the fibre, and 23 with the stress across it. That 
the plates were found stronger in a direction at right angles to 
that in which they were rolled is probably due to some error 
in marking the plates. 

" Making certain empirical allowances, Sir Wm. Fairbairn 
adopted the following ratios as expressing the relative strength 
of riveted joints : 



560 APPLIED MECHANICS. 

Solid plate 100 

Double-riveted joint 70 

Single-riveted joint 50 

These well-known ratios are quoted in most treatises on rivet- 
ing, and are still sometimes referred to as having a considerable 
authority. 

" It is singular, however, that Sir Wm. Fairbairn does not 
appear to have been aware that the proportion of metal 
punched out in the line of fracture ought to be different in 
properly designed double and single riveted joints. These 
celebrated ratios would therefore appear to rest on a very 
unsatisfactory analysis of the experiments on which they are 
based. Sir Wm. Fairbairn also gives a well-known table of 
standard dimensions for riveted joints. It is not very clear 
how this table has been computed, and it gives proportions 
which make the ratio of tearing to shearing area different for 
different thicknesses of plate. There is no good reason for 
this." 

As to the tests which constitute the experimental work of 
the committee, these were made by or under the direction of 
Prof. A. B. W. Kennedy, of London. Steel plates and steel 
rivets were used throughout, the steel containing about 0.18 per 
cent of carbon, and having a tensile strength varying from 
about 62000 to about 70000 pounds per square inch, and hence 
being a little harder than would correspond to our American 
ideas of what is suitable for use in steam-boilers. The greater 
portion of the work was performed by the use of a testing- 
machine of 100000 pounds capacity, and hence one which did 
not admit of testing wide strips with a sufficient number of 
rivets to correspond to the cases which occur in practice; 
indeed, only eighteen of the tests were made on such strips. 
Nevertheless, a brief summary of what was done will be given 
here, though some of the conclusions which he drew are al 



RI VE TED JOINTS. 5 6 1 



ready, and others are liable to be, proved untrue by tests of 
wide strips. The tests made by Prof. Kennedy up to 1885 
consisted of fourteen series numbered I to V, VA and VI to 
XIII, and covering 290 experiments, 64 on punched or drilled 
plates, 97 on joints, 44 on the tenacity of the plates used in 
the joints, 33 on the tenacity and shearing-resistance of the 
rivet-steel used in the joints, and the remaining 52 on various 
other matters. 

The first three series were upon the tenacity of the steel 
used, and showed it to be, as stated, from 62000 to 70000 pounds 
per square inch, with an ultimate elongation of 23 to 25 per 
cent in a gauged length of ten inches ; the tenacity of the 
rivet-steel being practically the same as that of the plates. 
The fourth series showed the shearing-strength of the rivet- 
steel to be about 55000 pounds per square inch when tested in 
one way, and 59000 pounds per square inch when tested in 
another way which corresponded, as Kennedy claims, better 
to the conditions of a rivet, though neither was by using a 
riveted joint. 

The tests of series V and VA were made upon pieces of 
plate which had been punched or drilled, in other words, on 
grooved specimens ; and, as might be expected, these specimens 
showed invariably an increase in tensile strength- over the 
straight specimens. In the J" and -J*" plates drilled with holes 
1 inch in diameter and 2 inches pitch, the net metal between 
the holes had a tenacity 11 to 12 per cent greater than that of 
the untouched plate. Even with punched-holes the metal had 
a similar excess of tenacity of over 6 per cent. The remaining 
eight series, VI to XIII inclusive, were made on riveted joints, 
the first five on single-riveted lap-joints, and the last three, 
or XI, XII, and XIII, on double-riveted lap and butt joints. 

Series VI was made on twelve joints in -f-inch plates which 
contained only two rivets each, the proportions not being in- 
tended to be those of practice, but such as should give, to 



562 APPLIED MECHANICS. 

some extent, limiting values for the resistances of the plate to 
tearing, and of the rivets to shearing and pressure. The results 
were rather irregular; and the main conclusion which he drew, 
was, that if the joint is not to break by shearing, the ratio of 
the tearing to the shearing area must be computed on a much 
lower value of shearing-strength per square inch than the ex- 
periments of series IV had shown ; indeed, some of the joints 
of series VI gave way by shearing the rivets at loads no greater 
than 36000 pounds per square inch of shearing-area. 

Series VII was made upon six (single-riveted lap) joints in 
f-inch plate, with only three f-inch rivets in each joint, and 
with varying pitch and lap ; all these joints breaking by shear- 
ing the rivets. His conclusion from these tests was, that the 
lap need not be more than 1.5 times the diameter of the rivet. 

Series VIII was made on eighteen (single-riveted lap) joints 
in six sets of three each, and these are the only single-riveted 
lap-joints which he tested, having as many as seven rivets each. 
The results are given in the accompanying table. 

Before giving the table, it may be said that No. 652 was in- 
tended to have such proportions as to be equally likely to give 
way by tearing or by shearing, the intensity of the shearing- 
strength being assumed as two-thirds that of the tensile 
strength of the steel, while the bearing-pressure per square 
inch was intended to be about 7.5 per cent greater than the 
tension. No. 653 was proportioned with excess of shearing or 
rivet-area, No. 654 with defect of shearing-area, No. 655 with 
excess of tearing or plate area, No. 656 with defect of tearing- 
area, and No. 657 with excess of bearing-pressure, the different 
pioportions being arrived at by varying the pitch and diameter 
of the rivets, and, in the case of 657, the thickness of the plate 
also. The margin (or lap minus radius of rivet) was J inch in 
each case. The following table will show how far these inten- 
tions were realized, and further comments will be deferred till 
later. 



RIVETED JOINTS. 



563 



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564 APPLIED MECHANICS. 

Series IX was made on twenty-one joints in f-inch plate 
(each containing only two rivets) designed in a manner similar 
to series VIII, while three were afterwards made from some 
of the broken plates, with as heavy rivets as it was deemed 
possible to make tight. 

From these tests Kennedy thinks it fair to conclude — 

1 °. That the efficiency of a single-riveted lap-joint in a j-inch 
plate cannot be greater than 50 per cent, unless rivets larger 
than I.I inch are used; and he also calls attention to the fact 
that, as he claims, strength is gained by putting more metal in 
the heads and ends of the rivets, claiming that it will make 
also a tighter joint for boiler-work. 

Series X was made on eight single-riveted lap-joints in 
J-inch and f-inch plate, made from the broken specimens of 
series V and VA ; they also had only two rivets each. These 
joints were made with a view of investigating the effect of 
more or less bearing-pressure. He claims that high bearing- 
pressure induces a low shearing-strength in the rivets, and that 
the bearing-pressure should not exceed about 96000 pounds per 
square inch ; also, that when a large bearing-pressure is used, 
the " margin " should be extra large to prevent distortion, and 
consequent local inequalities of stress ; also, that smaller bearing- 
pressures do not much affect the strength of the joint one way 
or the other. 

Series XI was made upon twelve specimens of double-riveted 
joints ; three being lap-joints in f-inch plate, three lap-joints in 
f-inch plate, three butt-joints with two equal covers in f-inch 
plate, and three butt-joints with two equal covers in f-inch 
plate. Kennedy designed these joints with a view to their 
being equally likely to fail by tearing or by shearing. His as- 
sumptions and the results of the tests are all given in the fol- 
lowing table : 



RIVETED JOINTS. 



565 



SERIES XI. DOUBLE-RIVETED LAP AND BUTT JOINTS— AVERAGES. 







■S a 

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LAP-JOINTS. 


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70560 


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2.15 


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75150 


53920 


91530 


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67200 


42560 


4.4 


3.18 


26 


100800 


59290 


37650 


88460 


71.3 



Series XII contains the same joints as series XI, the strained 
ends having been cut off, and the rest redrilled and riveted by 
means of Mr. Twedell's hydraulic riveter; and series XIII con- 
tained the same joints treated a second time in the same way. 
These experiments, so far as they went, showed no gain in 
ultimate strength to result from hydraulic as compared with 
hand-riveting ; but it was found that, through a misunderstand- 
ing, they had been riveted up at a pressure much lower than 
that intended by Mr. Twedell. 

On the other hand, the load at which visible slips occurred 
was about twice as much greater with hydraulic as with hand 
riveting. 



566 APPLIED MECHANICS. 



Kennedy's conclusions. 

The following are a portion of what he gives as his con- 
clusions : 

i°. The metal between the rivet-holes had a considerably 
greater tensile resistance per square inch than the unperf o* 
rated metal. 

2°. In single-riveted joints, with the metal that he used, he 
assumed about 22 tons (49280 lbs.) per square inch as the shear- 
ing-strength of the rivet-steel when the bearing-pressure is 
below 40 tons (89600 lbs.) per square inch. In double-riveted 
joints with rivets of about f-inch diameter we can generally 
assume 24 tons (53760 lbs.) per square inch, though some fell 
to 22 tons (49280 lbs.). 

3 . He advises large rivet heads and ends. 

4 . For ordinary joints the bearing-pressure should not ex- 
ceed 42 or 43 tons (94000 or 96000 lbs.) per square inch. For 
double-riveted butt-joints a higher bearing-pressure may be 
allowed ; the effect of a high bearing-pressure is to lower the 
shearing-strength of the steel rivets. 

5°. He advises for margin the diameter of the hole, except 
in double-riveted butt-joints, where it should be somewhat 
larger. 

6°. In a double-riveted butt-joint the net metal, measured 

zigzag, should be from 30 to 35 per cent greater than that meas- 

2 d 

ured straight across, i.e., the diagonal pitch should be —p -J — , 

where/ = transverse pitch and d ' = diameter of rivet-hole. 

7°. Visible slip occurs at a point far below the breaking* 
load, and in no way proportional to that load. 

Kennedy thinks that these tests enable him to deduce rules 
for proportioning riveted joints, and the following are his rules, 
viz.: 



RIVETED JOINTS, 567 



(a) For single-riveted lap-joints the diameter of the hole 
should be 2\ times the thickness of the plate, and the pitch of 
the rivets 2| times the diameter of the hole, the plate-area being 
thus 71 per cent of the rivet-area. If smaller rivets are used, 
as is generally the case, he recommends the use of the follow* 
ing formula : 

* d * a.; 

where / = thickness of plate, d = diameter of rivet, and p = 
pitch. 

For 30-ton (67200 lbs.) plate, and 22-ton (49280 lbs.) rivets, a = 0.524 
For 28-ton (62720 lbs.) plate, and 22-ton (49280 lbs.) rivets, a = 0.558 
For 30-ton (67200 lbs.) plate, and 24-ton (53760 lbs.) rivets, a == 0.570 
For 28-ton (62720 lbs.) plate, and 24 ton (53760 lbs.) rivets, a = 0.606 

Or, as a mean, a = 0.56. 

(J>) For double-riveted lap-joints he claims that it would be 
desirable to have the diameter of the rivet 2\ times the thick- 
ness of the plate, and that the ratio of pitch to diameter of 
hole should be 3.64 for 30-ton (67200 lbs.) plate and 22-ton 
(49280 lbs.) or 24-ton (53760 lbs.) rivets, and 3.82 for 28-ton 
(62720 lbs.) plate. 

Here, however, it is specially likely that this size of rivet 
may be inconveniently large, and then he says they should be 
made as large as possible, and the pitch should be determined 
from the formula to 

where, 

For 30-ton (67200 lbs.) plate, and 24-ton (53760 lbs.) rivets, a =: 1.16 
For 28-ton (62720 lbs.) plate, and 22-ton (49280 lbs.) rivets, a = 1.16 
For 30-ton (67200 lbs.) plate, and 22-ton (49280 lbs.) rivets, a = 1.06 
For 28-ton (62720 lbs.) plate, and 24-ton (53760 lbs.) rivets, a = 1.24 



568 APPLIED MECHANICS. 

(c) For double-riveted butt-joints he recommends that the 
diameter of the hole should be about 1.8 times the thickness 
of the plate, and the pitch 4.1 times the diameter of the hole, 
and that this latter ratio be maintained even when the former 
cannot be. 

Two of the principal participants in the discussion of the 
report were Mr. R. Charles Longridge and Prof. W. C. Unwin. 

Mr. Longridge was of the opinion that wider strips with 
more rivets should have been used ; that holding the specimens 
in the machine by means of a central pin at each end was not 
the best method ; that the results obtained from specimens 
which had been made from the remnants of other fractured 
specimens were at least questionable, for, even if the plate had 
not been injured, the ratio of the length to the width of the 
narrowest part was different after the strained ends were cut 
off from what it was before ; that machine-riveting should have 
been adopted throughout instead of hand-riveting, as it is not 
possible to secure uniformity with the latter even were it all 
done by the same man, as he would be more tired at one time 
than at another ; that experiments should be made to determine 
the effect of different sizes and different shapes of heads, as 
well as of different pressures upon the load causing visible slip , 
and that experiments should be made upon chain-riveting, as 
he thought the chain-riveted joint would show a greater effi- 
ciency than the staggered. 

Professor Unwin said : 

i°. In examining the results to ascertain how far a variation 
from the best proportions was likely to affect the strength of 
the joint, he found that while the ratio of rivet diameter to 
thickness of plate varied 21 per cent, the ratio of shearing to 
tearing area 30 per cent, and the ratio of crushing to tearing 
area 34 per cent, the efficiency of the weakest joint was only 
six per cent less than that of the strongest, or, in other words, 



RIVETED JOINTS. 569 



the whole variation of strength was only 1 1 per cent of the 
strength of the weakest joint. 

2°. With reference to the effect which the crushing-pres- 
sure on the rivet produced upon the strength of the joint, 
there were some old experiments, which showed that, when 
the bearing-pressure on the rivet became very large there was 
a great diminution in the apparent tenacity of the plate in 
the case of riveted joints in iron. Why should the crushing- 
pressure affect either the tenacity of the plate or the shearing 
resistance of the rivet? He believed that it did not really 
affect either. What happened was that, if the crushing-pres- 
sure exceeded a certain limit, there was a flow of the metal, 
and the section which was resisting the load was diminished. 
Either the section of the plate in front of the rivet, if the plate 
was soft, or the section of the rivet itself, if the rivet was soft, 
became reduced. 

3 . He thought that the point at which visible slip began 
was the initial point at which the friction of the plates was 
overcome, and of course was greater the greater the grip 
upon the plates, and hence greater in machine than in hand 
riveting. In some cases with hydraulic riveting loads were got 
as high as 10 tons (22400 lbs.) per square inch of rivet section 
before slipping began. 

4 . In regard to the rules for proportioning riveted joints, 
he preferred to distinguish the joints as single-shear and double- 
shear joints, and then we have the following three equations : 
one by equating the load to the tearing-resistance of the plates, 
a second by equating it to the shearing-resistance of the rivets, 
and a third by equating it to the crushing resistance ; these 
three determining the thickness of the] plate, the diameter of 
the rivet, and the pitch. 

By taking the crushing as double the tenacity, we should 
obtain for single shear d= 2.57/, and for double-shear, d = 
1.27/. 



57° APPLIED MECHANICS. 

In a single-shear joint the rivet cannot generally be made 
so big, and in the double-shear it could not always be made so- 
small, hence the rivet diameter is chosen arbitrarily, and then 
the single-shear joint is proportioned by the equations for shear- 
ing and tearing, no attention being paid to the crushing, while 
the double-shear joint is proportioned by the equations for 
crushing and tearing, no attention being paid to the shearing. 

5°. The general drift of the report was to advocate the use 
of larger rivets. Whether this could be done or not, he could 
not say. For lap-joints it would increase the strength, whereas 
for double-shear joints he was not sure that it would not be 
better to diminish the size of the rivet, and hence the crushing, 
pressure. 

This report has been given so fully because it emanates 
from a committee of the British Institution of Mechanical 
Engineers; but inasmuch as series VIII is the only one where 
wide strips were used, it seems to the writer that any conclu- 
sions which may be drawn from any of the other tests given 
in the report require confirmation by tests on wide strips with 
more rivets, before being accepted as true. 

Government Experiments. — The references to these experi- 
ments have been mentioned on page ooo. 

Those included in the first five of the volumes mentioned 
may be divided into three parts : — 

i°. Those contained in the first two Executive Documents 
mentioned above. 

2°. Those contained in the third and fourth. 



RIVETED JOINTS. $?I 



3°. Those contained in the fifth. 

Summaries of these sets of tests will be given here in their 
order, as each set was made with certain special objects in 
view, and, if not all, at any rate the i° and 2°, form, as has been al- 
ready stated, the first portion of a systematic series ; and it seems 
to the author that, although the series are not yet completed, 
yet these tests themselves furnish more reliable information in 
regard to the behavior and the strength of joints than any other 
experiments that have been made, and that the figures them- 
selves furnish the engineer with the means of using his judg- 
ment in many cases where he had no reliable data before. 

A perusal of the tables will give a good idea of the shear- 
ing-strength per square inch of the rivet iron, which is seen to 
be less than the tensile strength of the solid plate ; also the 
effect on strength of the plates due to the entire process of 
riveting, punching, drilling, and driving the rivets ; also the 
efficiencies of the joints tested. 

One of the strongest single-riveted joints tested was a single- 
riveted lap-joint with a single covering-strip. 

The apparent anomaly of the punched plates in a few cases, 
showing a greater strength than the drilled plates, is explained 
by Mr. Howard to be due to the strengthening effect of cold- 
punching combined with smallness of pitch, inasmuch as then 
the masses of hardened metal on the two sides re-enforce each 
other. 

Further than this, the student is left to study the figures 
themselves as to the effect of different proportions, etc. 

In regard to the first series, i.e., those contained in the first 
two Executive Documents mentioned, it is stated in the report 
that — 

i°. " The wrought-iron plate was furnished by one maker 
out of one quality of stock." 

2° " The steel plates were supplied from one heat, cast in 
ingots of the same size; the thin plates differing from the 



$72 APPLIED MECHANICS. 

thicker plates only in the amount of reduction given by the 
rolls." 

The modulus of elasticity of the metal was, iron plate 
31970000 lbs. ; steel plate, 28570000 lbs. 

In the tabulated results, the manner of fracture is shown 
by sketches of the joints, and is further indicated by heavy 
figures in columns headed " Maximum Strains on Joints, in ibs., 
per Square Inch." 



RIVETED JOINTS. 



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Tensile 
Strength, 

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.f, <u 


»o "o 


"C-3 


t^'nx 


"S J5 


"C .c 


ex 


^ rC 


•S * H - fi 


^ J3 


u ja 


^ J2 


l! k 


CJ 


CJ V 


_ (_) 


CJ 


_ o 


^-r O 


•^ ° •— u 


rt o 


— •• u 


— CJ 


c c 


3 3 3 c 


c c 


c c 


S c 


^ C 


4) C <U C 


4) C 


4) C 


S^ 3 


W-a 


p 3 


O 3 O 3 


2 s 


p 3 


P 3 


£> 2 


_4j 3 4) 3 


« 3 


*> 3 

CO P-. 


Si 3 


•a £ 

u 

.a 

co 


.fc! d. 


K CU3 d 


.^ Cl, 


.fe Ph 


™ Pi 


CO On 


"So Onto C-. 


to O, 


CO Oh 


.S.S 


.5 .S.S .5 


.S.S 


•S.S 


.S.S 


.S.S* 


.S.S .S.S 


c c 


.S.S 


.s .s 


£*i 


«f*2fe2P ' 


lCf£> 


^"H 


:£■** 


•CfO 




II 


-I'ocdf 


a 






1-1 
















<^J1 


S v £ 

rt 5J2 


13 


13 13 


13 


13 


13 


13 


13 13 


13 


13 


13 


V 


4) 4> 


1) 


<v 


i> 


CU 


4) 4) 


4> 


4> 


cu 


co 


CO CO 


CO 


CO 


co 


CO 


CO CO 


to 


CO 


CO 


3ho 


.3 


.S .£ 

Kpo -p» 


.5 

ecf» 


.s 




~l» 


.s .s 


.S 
>qoo 


.s 


3 


c 
'o 




i-'S 










«"T5 








1— J 




.5 > '- 










r-- 4) *J 








"a 












Il| 








V 

'B 

CO 




00 '£ 










C7TE 




























' 




O 


r-, o 




VO 


r^ 


O 


•H ^> 


vo 


00 


On 


•JS3X JO -0 N 


N 


N CO 


CO 


VO 


VO 


!>. 


r^. r^ 


r-^ 


t^ 


r^ 


tJ- 


^ Tf 


"«*■ 


r>. 


t^ 


t^ 


x^ r^ 


t^ 


1^ 


r^ 






-<*- 


"* "* 


Tj- 

















RIVETED JOINTS. 



593 



GOVERNMENT TESTS OF GROOVED SPECIMENS. 



Tensile Test 


* of i-in. 


Grooved Specimens 


Wrought 


Iron 


Punched. 


S 




Sjs 












c a 


o . 




8 


M « 









</) 


m rf, 


OS o 

^6 


c v 


DM t/i 


JS* 


IS* 


3 a-S 


^ 


H 


& 


Inch. 


Inch. 




0.48 


0.240 


48090 


0.46 


0.235 


46940 


0.46 


0.241 


49280 


0.49 


0.240 


5534o 


0.44 


0.239 


5!52o 


0.47 


0.241 


49910 


0.97 


0.247 


49540 


0.98 


0.247 


49960 


0.94 


0.249 


50128 


0.96 


0.248 


46900 


0.98 


0.250 


46980 


0.96 


0.251 


46350 


1.47 


0.250 


37636 


1.50 


0.252 


37326 


1.48 


0.249 


41030 


1.48 


0.247 


39480 


i-47 


0.250 


37446 


i-45 


0.251 


39533 


1.96 


0.281 


43194 


i-95 


0.274 


4749° 


i-95 


0.282 


41360 


1.92 


0.279 


43080 


2.03 


0.250 


41140 


1.99 


0.248 


39575 


2.42 


0.280 


36210 


2.40 


0.245 


42245 


2.47 


0.243 


42233 


2.46 


0.285 


42712 


2.48 


0.245 


38125 


2.44 


0.248 


41620 


2.97 


0.247 


38964 


2.98 


0.241 


41540 


2.96 


0.241 


39972 


2.92 


0.240 


41712 


2.98 


0.250 


40430 


a-95 


0.247 


40850 



Tensile Tests of 4-fn. 


Grooved Specimens 


Wroueht-Iron 


Drilled. 


£ 




■Sjz 












«J3 








m « 









w 


Ui a 4 


rt O 


c 6 


VUi </,* 

Us 


is 


H 





Inch. 


Inch. 




0.51 


0.249 


55787 


0.52 


0.245 


55905 


0.52 


0.275 


5748o 


0.52 


0.276 


56000 


0.49 


0.248 


49600 


0.50 


0.248 


56700 


0.47 


0.275 


54880 


0.51 


0.276 


57800 


1. 00 


0.276 


54300 


1.02 


0.273 


57700 


1. 00 


0.276 


53800 


1. 00 


0.280 


52430 


1. 00 


0.252 


49400 


1.02 


0.275 


54060 


1. 01 


0.247 


52770 


1. 00 


0.278 


54600 


1.50 


0.276 


49130 


1.52 


0.273 


51300 


1.48 


0.251 


47220 


1.51 


0.273 


53400 


i-5 2 


0.275 


54180 


1.50 


276 


54600 


1.48 


0.274 


56250 


1.50 


0.249 


46260 


2.01 


0.275 


45900 


2.05 


0.279 


46820 


2.00 


0.275 


47950 


2.00 


0.278 


49640 


2.00 


0.286 


44650 


2.00 


0.275 


50780 


2.02 


0.279 


48850 


2.00 


0.277 


49840 


2-51 


0.244 


44980 


2.52 


0.280 


40150 


2.51 


0.282 


43150 


2.50 


0.244 


45500 


2.51 


0.285 


46500 


2.49 


0.242 


49520 


2.49 


0.242 


— 


2.50 


0.280 


44780 


3.02 


0.250 


45700 


3.02 


0.249 


44870 


3.00 


0.240 


46760 


3.00 


0.250 


45700 


2-93 


0.242 


47950 


2.99 


0.250 


48740 


2.98 


0.279 


45900 


3.01 


0.281 


44410 



Tensile Tests of i-in. 


Grooved Sp 
Steel PI 


ecimens 


ate 


Punched. 


e 




•£^= 













c c 


„: 






W g 





£ 


S2 


V 


w 6< 


xO 




S ^~ 


W 


•Js£ 


S^.5 


& 


H 


13 


Inch. 


Inch. 




o.49 


0.250 


65120 


0.47 


0.249 


67010 


0.48 


0.249 


63420 


0.48 


0.248 


66550 


0.48 


0.247 


67060 


0.47 


248 


65300 


0.99 


0.249 


59840 


1. 00 


0.250 


62160 


1. 01 


0.249 


68246 


0.96 


0.250 


67330 


0.96 


0.248 


65966 


o.95 


0.245 


62700 


i-45 


0.248 


64080 


i-45 


0.252 


64000 


i-45 


0.249 


61025 


i-Si 


0.251 


59420 


1.96 


0.250 


59900 


1-93 


0.252 


63500 


1.98 


0.250 


59350 


1.96 


0.251 


59060 


2.49 


0.249 


58100 


2.47 


0.249 


63900 


2-43 


0.250 


61640 


2-95 


0.251 


56530 


3.01 


0.249 


58780 


3-04 


0.253 


55500 


2.97 


0.252 


60060 


2.98 


0.251 


54050 


2.97 


0.249 


56040 



1 

Tensile Tests of i-in. 


Grooved Specimens 


Steel Plate 


Drilled. 


6 




■5 .a 














„; 






m £ 







*j 


w 


w 6- 


M O 


v • 


VUi ui 


^O 




6 * rt 


P 


£P4 


2 a. 2 


h 


£ 


Inch. 


Inch. 




0.52 


0.246 


67890 


0-54 


0.248 


67160 


o.53 


0.247 


66870 


0.50 


0.247 


65610 


0.51 


0.249 


66370 


0.51 


0.250 


67420 


0.52 


0.248 


67750 


0.52 


0.252 


61910 


1.03 


0.247 


67090 


1.02 


0.250 


66390 


1.02 


0.246 


66770 


1.02 


0.250 


67730 


I.OI 


0.247 


66020 


1. 00 


0.251 


67010 


1. 00 


0.247 


64450 


I.OI 


0.250 


66090 


1.54 


0.250 


64390 


1.52 


0.251 


63350 


1.50 


0.253 


64370 


i-54 


0.248 


64895 


2.02 


0.252 


64320 


2.00 


0.251 


62970 


2.00 


0.251 


60910 


2.50 


0.248 


59260 


2.50 


0.252 


63250 


2-53 


0.248 


59390 


3-03 


0.251 


61577 


3.00 


0.249 


59080 


3.02 


0.251 


59550 


3.02 


0.250 


59700 


3.00 


250 


63370 


3.00 


0.251 


58630 ■ 


3-03 


0.252 


63940 ; 



594 



APPLIED MECHANICS. 



Iron Ponched. 



Iron Drilled. 



Steel Punched. 



Steel Drilled. 



Tensile Tests of 

Grooved Wrought- 

Iron Plates. 


I 

M 

% 


u5 

V 

a 

M 

M 

H 


3 m 

Hen w 


Inch. 


Inch. 




I.OI 


o-373 


47000 


0.98 


0.370 


47520 


2.00 


0.382 


39760 


2.02 


0.383 


36630 


2. 3 Q 


0.390 


37600 


2.98 


o-395 


36340 


2.98 


0.392 


39210 


3-47 


0.390 


37680 


3-47 


0.389 


38340 


0.97 


0.467 


50820 


1.48 


0.506 


45090 


1.49 


0.506 


45050 


1.91 


o-5i3 


42500 


1.97 


0.512 


43430 


2.47 


0.516 


394IO 


2.41 


0.513 


39720 


3.00 


0.515 


38950 


2.90 


0.517 


37290 


3-5° 


0.520 


37800 


3-49 


0.513 


37770 


4.00 


0.515 


35730 


4-03 


0.516 


36690 


3-99 


0.511 


37000 


4-°3 


0.508 


37420 


0.97 


0.614 


49770 


I.OI 


0.619 


52960 


1.48 


0.618 


46320 


i-5 2 


0.620 


46750 


2.99 


0.614 


40140 


3-5° 


0.615 


3748o 


3-5° 


0.616 


36940 


4.04 


0.619 


373IO 


0.98 


0.678 


50840 


1 I.OI 


0.682 


46590 


1 i-49 


0.688 


45970 


i 3-48 


0.691 


40350 


3-53 


0.692 


39380 



Tensile Tests of 


Grooved Wrought- 
Iron Plates. 


M 
•73 


(/) 

<D 

c 


M 

H 


3 m 

lis 

{3 


Inch. 


Inch. 




0.98 


0.376 


50870 


0.98 


o.377 


52660 


1.98 


o.379 


49710 


2.00 


0.380 


49830 


2.50 


0.390 


50250 


3.00 


0.392 


45!5o 


3.00 


0-393 


47540 


3-50 


0.392 


43940 


3-49 


0.390 


46490 


0.99 


0.477 


47140 


1. 00 


o.479 


48370 


1.49 


0.510 


51240 


1.49 


0.512 


51510 


1.98 


0.514 


50050 


1.98 


0.516 


47790 


2.51 


0.520 


4558o 


2.52 


0.516 


44960 


3.00 


0.515 


44980 


3.01 


0.519 


47030 


3-51 


0.513 


46170 


3-49 


0.514 


44760 


3-99 


0.510 


45330 


3-98 


0.513 


45000 


4.00 


0.506 


46100 


o.97 


0.628 


47220 


1. 00 


0.626 


48350 


1.52 


0.625 


47170 


1.49 


0.629 


4653° 


2.98 


0.613 


48220 


3-40 


0.616 


4777° 


3-47 


0.617 


44900 


3-9 1 


0.625 


44840 


3-9° 


0.626 


45ioo 


o.99 


0.695 


47500 


0.99 


0.691 


52780 


I-5I 


0.692 


48470 


3-44 


0.700 


4775o 


3-49 


0.692 


46350 



Tensile Tests 




of 




Grooved Steel Plates. 






3 m 












a 






















in 


w 6- 




X 






M 


s a 




I 


IS.S 
5 


Inch. 


Inch. 




1.99 


0.365 


61890 


0.99 


0.494 


70080 


1. 00 


0.492 


68130 


1.50 


o.497 


66340 


1.51 


0.494 


63810 


1.99 


0.499 


55930 


1.97 


0.500 


64260 


2.43 


0.502 


52050 


2.51 


0.504 


64360 


3.00 


0.503 


60320 


2.99 


0.503 


62430 


3-50 


0.503 


49430 


3-50 


0.505 


48270 


4.00 


0.497 


48010 


4.00 


0.499 


55190 


3-99 


0.501 


55780 


3-99 


0.498 


46250 


I.OI 


0.613 


66720 


1-52 


0.612 


64800 


i-5° 


0.615 


64400 


2.50 


0.618 


58060 


2.52 


0.619 


58780 


2.99 


0.617 


57180 


3.46 


0.615 


58410 


3-5i 


0.615 


57 J 9o 


4.04 


0.612 


5445o 


4-03 


0.614 


5738o 


I.OI 


0.721 


67930 


1. 00 


0.718 


67620 


1.50 


0.719 


62890 


3-50 


o.735 


56730 


3-5i 

._ 


0-733 


54220 



Tensile Tests 

of 

Grooved Steel Plates. 


•3 


tn 

u 

c 

M 

H 


3m 

Mo 

c c 

u 

« - Q 

3 S.S 

13 


Inch. 


Inch. 




1.97 


0.369 


63620 


1. 00 


0.498 


66220 


0.99 


o.495 


•66800 


1. 00 


0.500 


67000 


i-53 


0.497 


65930 


1.50 


0.498 


66270 


1.98 


0.504 


67510 


2.03 


0.502 


66730 


2.50 


0.497 


67950 


2.52 


0.501 


67440 


3.01 


0.502 


66310 


3.01 


0.503 


66190 


3-49 


0.504 


64920 


3-50 


0.502 


65210 


3-99 


0.499 


64470 


4.00 


0.498 


64810 


4.00 


0.503 


64690 


4.00 


0.498 


64140 


0.99 


0.619 


60290 


1.49 


0.614 


63610 


1.49 


0.616 


63450 


2-49 


0.620 


59*70 


2.50 


0.619 


59600 


3-oi 


0.617 


59270 


3-50 


0.614 


61610 


3-49 


0.617 


62060 


4.00 


0.615 


60330 


4.01 


0.617 


61 1 20 


0.96 


0.726 


58480 


I.OI 


0.727 


58790 


1-5* 


0.726 


59290 


3-50 


0.736 


58700 


3-49 


0.729 


59180 



TENSILE TESTS OF RIVETED JOINTS. 595 



Next will be given the two series of tests already referred 
to, with Mr. Howard's analysis of them. 

TENSILE TESTS OF RIVETED JOINTS. 

" Earlier experiments on this subject made with single and 
double riveted lap and butt joints in different thicknesses of 
iron and steel plate, together with the tests of specimens pre- 
pared to illustrate the strength of constituent parts of joints, 
are recorded in the report of tests for 1882 and 1883. 

From the results thus obtained it appeared desirable to 
institute a synthetical series of tests, beginning with the most 
elementary forms of joints in which the stresses are found in 
their least complicated state. To meet these conditions, a 
series of joints have been prepared which maybe designated as 
single-riveted butt-joints, in which the- covers are extended so 
as to be grasped in the testing-machine ; thereby enabling one 
plate of the joint to be dispensed with, and securing the test of 
one line of riveting. 

Such a joint, made with carefully annealed mild steel plate 
of superior quality, with drilled holes, seems well adapted to 
demonstrate the influence on the tensile strength of the metal 
taken across the line of riveting, of variations in the width of 
the net section between rivets, and variations in the compres- 
sion stress on the bearing-surface of the rivets ; elements which 
are believed to be fundamental in all riveted construction. 

This series comprises 216 specimen joints, the thickness of the 
plate ranging from J" to |", advancing by eighths. The covers 
are from -^" to ^". The rivets are wrought-iron, and range from 
■fc" to iyV' diameter ; they are machine-driven in drilled holes 
J-g-" larger in diameter than the nominal size of the rivets. Ten- 
sile tests of the material accompany the tests of the joints. 

From each sheet of steel two test-strips were sheared, one 
lerf^thwise and one crosswise. The strips were 2\" wide and 24" 
to 36" long; they were annealed with the specimen plates,and had 
their edges planed, reducing their widths to \\" before testing. 



596 APPLIED MECHANICS. 

Micrometer readings were taken in 10" along the middle of 
the length of each. 

The strength and ductility appear to be substantially the 
same in each direction. But the practice of the rolling-mill 
where these sheets were rolled is such that nearly the same 
amount of work may have been given the steel in each direc- 
tion ; that is, lengthwise and crosswise the finished sheet. 

The ingots of open-hearth metal are first rolled down to 
slabs about 6" thick, then . reheated and rolled either length- 
wise or crosswise their former direction, as best suits the re- 
quired finished dimensions. 

The tensile tests show among the thinner plates a relatively 
high elastic limit as compared with the tensile strength ; in the 
■fo" plate the percentage is 72.2, while with the f /7 plate the 
percentage is found to be 53.3. 

It is noticeable that the thinner plates particularly exhibit 
a large stretch immediately following the elastic limit, and the 
stretching is continued at times under a load lower than that 
which has been previously sustained. It is characteristic of all 
the thicknesses that a considerable stretch takes place under 
loads approaching the tensile strength — in some cases the 
stretch increases 5 to 6 per cent, while the stress advances 1000 
pounds per square inch or less. Herein is found a valuable 
property of this metal as a material for riveted construction. 
The stress from the bearing-surface of the rivets is distributed 
over the net section of plate between the rivets, due to the large 
stretch of the metal, with little elevation of the stress, and a 
nearer approximation of uniform stress in this section attained 
than is found in a brittle or less ductile metal. The joints were 
held for testing in the hydraulic jaws of the testing-machine, 
having 24" exposure between them. A loose piece of steel 
the same thickness as the plate was placed between the covers 
to receive the grip of the jaws, and avoid bending the covers. 

Elongations were measured in a gauged length of 5", the 
micrometer covering the joint at the middle of its width. Loads 



TENSILE TESTS OF RIVETED JOINTS. 597 

were applied in increments of 1000 pounds per square inch of the 
gross section of the plate, the effect of each increment determined 
by the micrometer, and permanent sets observed at intervals. 

The progress of the test of a joint is generally marked 
by three well-defined periods. In the first period greatest 
rigidity is found, and it is thought that the joint is now held 
entirely by the friction of the rivet-heads, and the movement of 
the joint is principally that due to the elasticity of the metal. 

The second period is distinguished by a rapid increase in 
the stretch of the joint ; attributed to the overcoming of the 
friction under the rivet-heads and closing up any clearance 
about the rivets, bringing them into bearing condition against 
the fronts of the rivet-holes. Rivets which are said to fill the 
holes can hardly do so completely, on account of the contrac- 
tion of the metal of the rivet from a higher temperature than 
that of the plate, after the rivet is driven. 

After a brief interval the movement of the joint is retarded, 
and the third period is reached. The stretch of the joint is 
now believed to be due to the distortion of the rivet-holes and 
the rivets themselves. The movement begins slowly, and so 
continues till the elastic limit of the metal about the rivet-holes 
is passed, and general flow takes place over the entire cross-sec- 
tion, and rupture is reached. These stages in the test of a joint 
are well defined, except when the plates are in a warped condi- 
tion initially, when abnormal micrometer readings are observed. 

The difference in behavior of a joint and the solid metal 
suggests the propriety of arranging tension joints in boiler con- 
struction and elsewhere as nearly in line as practicable. 

The efficiencies of the joints are computed on the basis of 
the tensile strength of the lengthwise strips, this being the 
direction in which the metal of the joints is strained. The 
efficiencies here found are undoubtedly lowered somewhat by 
the contraction in width of the specimens, causing in most cases 
fractures to begin at the edges and extend towards the middle 
of the joint. Of the entire series, 88 joints have been tested ; 
tfc" i"> f"> an d f " plates yet remain." 



59« 



APPLIED MECHANICS. 



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TABULATION OF 0. H. STEEL STRIPS. 



599 



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6oo 



APPLIED MECHANICS. 



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RIVET METAL FOR RIVETED JOINTS. 



601 







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602 



APPLIED MECHANICS. 



TABULATION OF SINGLE- 

i" STEEL 



No. of 
Test. 

1308 


Sheet Letters. 


Pitch. 


No. of 
Rivets. 


Width 

of 
Joint. 


Nominal 
Thickness. 


Size of 
Rivets 

and 
Holes. 


Actual 
Thick- 
ness of 
Plate. 


Lap. 


Plate. 


Covers. 


Plate. 


Covers. 


F 


A 


A 


in. 
if 


6 


in. 

9-75 


in. 

1/4 


in. 
3/16 


in. 

A*i 


in. 

.242 


in. 

2 


1309 


F 


A 


A 


" 


6 


14 


44 




" 


.242 


2 


1310 


F 


A 


A 


1* 


6 


10.50 


44 




ti 


.242 


2 


1311 


F 


A 


A 


" 


6 


44 


tt 




44 


.249 


2 


1312 


F 


A 


A 


if 


6 


11.25 


44 




ti 


.244 


2 


1313 


F 


A 


A 


" 


6 


" 


tt 




tt 


•243 


2 


13*4 


F 


A 


A 


1* 


6 


10.49 


44 




tt** 


.248 


2 


1315 


F 


A 


A 


44 


6 


it 


tt 




44 


.242 


2 


1316 


F 


A 


A 


»1 


6 


11.27 


44 




44 


.244 


2 


1317 


F 


A 


A 


44 


6 


44 


44 




44 


.246 


2 


1318 


F 


B 


B 


2 


6 


12.01 


tt 




44 


•245 


2 


1319 


G 


B 


B 


" 


6 


44 


44 




44 


.240 


2 


1320 


G 


B 


B 


2* 


6 


12.76 


" 




tt 


243 


2 


1321 


G 


B 


B 


" 


6 


" 


tt 




44 


•243 


2 


1322 


H 


D 


D 


2* 


6 


I3-5I 


'« 




44 


•245 


2 


1323 


H 


D 


C 


" 


6 


" 


44 




it 


.247 


2 


1324 


F 


A 


A 


if 


6 


11.26 


it 




tt*i 


.248 


2 


1325 


F 


A 


A 


" 


6 


" 


it 




44 


.245 


2 


1326 


G 


B 


B 


2 


6 


12.00 


44 




<• 


.241 


2 


1327 


G 


B 


B 


" 


6 


44 


44 




44 


.242 


2 


1328 


G 


B 


C 


2* 


6 


12.76 


44 




44 


.241 


2 


1329 


G 


C 


C 


" 


6 


44 


44 




44 


.242 


2 


'330 


L 


C 


D 


2* 


A 


i3-5o 


44 




44 


.248 


2 


1331 


H 


C 


D 


" 


6 


" 


tt 




44 


.246 


2 


J 332 


H 


D 


E 


2f 


6 


14.25 






44 


.248 


2 


*333 


H 


D 


E 


" 


6 


it 


it 




44 


.248 


2 


1334 


M 


E 


E 


=£ 


6 


15.00 


44 




44 


• 243 


2 


1335 


O 






" 


6 


ti 


44 




44 


•245 


2 


1336 


H 


C 


C 


2f 


5 


13-13 


<i 




tt 


238 


2 


1337 


L 


C 


C 


" 


5 


44 


44 




44 


.252 


2 


X338 


G 


B 


B 


2 


6 


12.00 


44 




tt*' 


.238 


2 


1339 


F 


B 


B 


44 


6 


44 


tt 




it 


.248 


2 


1340 


G 


B 


C 


a* 


6 


12.75 


44 




" 


.240 


2 


!34i 


G 


C 


C 


it 


6 


44 


44 




H 


.242 


2 


*342 


L 


C 


D 


4 


6 


i3-5i 


44 




ti 


.250 


2 


1343 


L 


D 


D 


44 


6 


1 1 


" 




II 


.250 


2 



TABULATION OF RIVETED JOINTS. 



603 



RIVETED BUTT-JOINTS. 
PLATE. 



Sectional Area 
of Plate. 


Bearing 
Surface 

of 
Rivets. 


Shear- 
ing 
Area of 
Rivets. 


Tensile 
Strength 
of Plate 

per 
Sq. In. 


Maximum Stress on Joint per 


Sq. In. 


Effi- 
ciency 

of 
Joint. 


Tension 

on Gross 

Section of 

Plate. 


Tension 

on Net. 

Section of 

Plate. 


Comp. on 
Bearing 
Surface 

of Rivets. 


Shear- 
ing of 
Rivets. 


Gross. 


Net. 


sq. in. 
2.360 


sq. in. 
1.452 


sq in. 

.908 


sq. in. 
3.682 


lbs. 
61740 


lbs. 
41690 


lbs. 
67 7 70 


lbs. 
108370 


lbs. 
26720 


67.5 


2.360 


1.452 


.908 


3.682 


61740 


42180 


68560 


109640 


27040 


68.3 


2 541 


1.634 


.907 


3.682 


61740 


42540 


66160 


119180 


29360 


68.9 


2.615 


1. 681 


•934 


3.682 


61740 


43170 


67160 


119810 


30660 


69.9 


2-745 


1.830 


•915 


3.682 


61740 


44920 


67380 


J 34750 


3349o 


72.8 


2-739 


1.827 


.912 


3.682 


61740 


44520 


66750 


133720 


33120 


72.1 


2.602 


1.486 


1.1x6 


5.300 


61740 


40700 


71270 


94890 


19980 


65-9 


2.541 


1.452 


1.089 


5-300 


61740 


40000 


70000 


93330 


19180 


64.8 


2.750 


1.652 


1.098 


5-3oo 


61740 


40980 


68210 


102630 


21260 


66.4 


2.772 


1.665 


T.107 


5.300 


61740 


4M3o 


68980 


103750 


21670 


67.1 


2.942 


1.840 


1. 102 


5-3oo 


61740 


42180 


67450 


112610 


23420 


68.3 


2.882 


1.802 


1.080 


5.300 


62660 


43000 


68770 


"4750 


23380 


68.6 


3.100 


2.007 


1.093 


5-300 


62660 


43060 


66520 


122140 


25190 


68.7 


3.100 


2.007 


1.093 


5.300 


62660 


44030 


68000 


124880 


2575o 


70.3 


3-3io 


2.207 


1. 103 


5.300 


59180 


43040 


64540 


129150 


26880 


72.7 


3-337 


2.225 


1. 112 


5.300 


59180 


43810 


65700 


131460 


27580 


74.0 


2.792 


1.490 


1.302 


7.216 


61740 


38650 


72480 


82870 


i495o 


62.0 


2.756 


1.470 


1.286 


7.216 


61740 


3943o 


73930 


84510 


15010 


63.8 


2.892 


1.627 


1.265 


7.216 


62660 


40340 


71700 


92210 


161 70 


64.4 


2.909 


1.638 


1. 271 


7.216 


62660 


41280 


73310 


94480 


16640 


65-9 


3-075 


1. 810 


1.265 


7.216 


62660 


42290 


71850 


102810 


18020 


67.5 


3.088 


1.817 


1. 271 


7.216 


62660 


42750 


72650 


103860 


18290 


68.2 


3-348 


2.046 


1.302 


7.216 


61470 


43100 


70530 


1 10830 


20000 


70.1 


3-321 


2.029 


1. 291 


7.216 


59180 


4!450 


67840 


106620 


19080 


70.0 


3 • 534 


2.232 


1.302 


7.216 


59180 


41820 


66210 


113500 


20480 


70.7 


' 3-534 


2.232 


1.302 


7.216 


59180 


42760 


67710 


1 16070 


20940 


72-3 


3 645 


2.369 


1.276 


7.216 


58170 


44650 


68700 


127550 


22550 


76.8 


3- 6 75 


2.389 


1.286 


7.216 


64170 


43050 


66230 


123030 


21930 


67.1 


3-125 


2.084 


1. 041 


6.013 


59180 


41310 


61960 


124020 


21470 


69.8 


3 300 


2.206 


1.103 


6.013 


61470 


42000 


62990 


125990 


23110 


68.3 


2.856 


1.428 


1.428 


9-425 


62660 


40620 


81230 


81230 


12310 


64.8 


2.976 


1.488 


1.488 


9-425 


61740 


36290 


72580 


72580 


1 1 460 


58.8 


3 .060 


1.620 


1.440 


9-425 


62660 


38660 


73020 


82150 


!2550 


61 .7 


3.088 


1.636 


1.452 


9-425 


62660 


38000 


71730 


80820 


12450 


60.6 


3-378 


1.878 


1.500 


9-425 


61470 


37800 


68000 


85130 


13550 


61.5 


3-375 


1-875 


1.500 


9-425 


61470 


39000 


70200 


87750 


13970 


63-4 



604 



APPLIED MECHANICS. 



TABULATION OF SINGLE- 
i" STEEL 



No. 

of 

Test. 


Sheet Letters. 


Pitch. 


No. 

of 

Rivets. 


Width 

of 
Joint. 


Nominal Thick- 
ness. 


Size of 
Rivets 

and 
Holes. 


Actual 
Thick- 
ness of 
Plate. 


Lap. 


Plate. 


Covers. 


Plate. 


Covers. 










in. 




in. 


in. 


in. 


in. 


in. 


in. 


1344 


L 


D 


E 


2f 


6 


14.24 


1/4 


3/16 


lf*i 


.250 


2 


1345 


H 


D 


E 


" 


6 


" 




" 


44 


it 


.247 


2 


1346 


I 


E 


E 


2* 


6 


15.00 




" 


44 


44 


.251 


2 


J 347 


I 


E 


E 


" 


6 


" 




' 


44 


44 


.252 


2 


1348 


H 


C 


C 


2| 


5 


13-13 




4 


44 


« 


.244 


2 


1349 


G 


C 


C 


" 


5 


" 




4 


tt 


«i 


•239 


2 


1350 


N 


D 


D 


2| 


5 


13-77 




4 


tl 


tt 


.250 


2 


i35i 


N 


D 


D 


" 


5 


" 




' 


u 


44 


.249 


2 


1352 


H 


D 


D 


2f 


5 


14-39 




1 


(< 


tc 


• 247 


2 


1353 


H 


E 


E 


ft 


5 


" 




4 


« 


44 


.248 


2 


*354 


I 


E 


E 


3 


5 


15.00 




4 


tc 


44 


.252 


2 


1355* 


1 


E 


E 


" 


5 


44 


" 


(( 


44 


.251 


2 



f" STEEL 



1356 


A 






if 


6 


9-75 


3/8 


1/4 


T%*§ 


•365 


2 


1357 


A 






41 


6 




44 




44 


•364 


2 


1358 


A 






i£ 


6 


10.49 


44 




44 


.365 


2 


1359 


A 






44 


6 




44 




" 


.366 


2 


1360 


A 






ii 


6 


10.50 


44 




H&i 


.366 


2 


1361 


A 






(C 


6 




44 




IC 


• 367 


2 


1362 


A 






I* 


6 


".25 


44 




44 


.366 


2 


1363 


A 






" 


6 




(t 




44 


• 365- 


2 


1364 


B 


K 


K 


2 


6 


12.00 


ti 




44 


.388 


2 


1365 


B 


K 


K 


44 


6 




44 




44 


•390 


2 


1366 


C 


K 


K 


2* 


6 


12. 7 6 


<( 




44 


•367 


2 f 


1367 


B 


K 


L 


44 


6 




44 




l( 


•387 


2 


1368 


A 


J 


J 


if 


6 


".25 


«« 




it&S 


•369 


2 


1369 


A 


J 


J 


44 


6 




44 




" 


.366 


2 


1370 


B 


K 


K 


2 


6 


12.00 


44 




44 


• 389 


2 


T 37i 


B 


J 


J 


" 


6 




tt 




it 


.388 


2 


I37 2 


B 


L 


L 


2* 


6 


12.77 


44 




44 


• 385 


2 


1373 


C 


K 


L 


44 


6 


44 


44 




44 


•367 


2 


1374 


D 


H 


N 


2} 


6 


13.50 


44 




44 


•376 


2 


1375 


E 


N 


N 


" 


6 




44 




44 


.380 


2 


1376 


D 


L 


M 


2f 


6 


14.23 


44 




tt 


•383 


2 


1377 


H 


L 


M 


44 


6 


44 






11 


•371 


2 



* Fractured two outside sections of plate at each edge along line 



TABULATION OF RIVETED JOINTS 



605 



RIVETED BUTT-JOINTS— Continued. 
PLATE— Continued. 



Sectional Area 


Bearing 
Surface 

of 
Rivets. 


Shear- 
ing 
Area of 
Rivets. 


Tensile 
Strength 
of Plate 

per 
Sq. In. 


Maximum Stress on Joint per Sq. In. 


Effi- 
ciency 

of 
Joint. 


of Plate. 


Tension 

on Gross 

Section of 

Plate. 


Tension 

on Net 

Sectionof 

Plate. 


Comp. on 
Bearing 
Surface 

of Rivets. 


Shear- 
ing of 
Rivets. 


Gross. 


Net. 


sq. in. 


sq. in. 


sq. in. 


sq. in. 


lbs. 


lbs. 


lbs. 


lbs. 


lbs. 




3-56o 


2.060 


1.500 


9-425 


61470 


39130 


67640 


92870 


14780 


63.6 


3-520 


2.038 


1.482 


9-425 


59180 


40450 


69860 


96070 


15110 


68.3 


3-765 


2.259 


1.506 


9-425 


60480 


4359o 


72640 


108960 


17410 


72.1 


3.780 


2.268 


1. 512 


9-425 


60480 


41420 


69030 


103540 


16610 


68.5 


3.204 


1.984 


1.220 


7-854 


59180 


38700 


62490 


101620 


15790 


65-4 


3-i3i 


1.936 


1 -195 


7-854 


62660 


42890 


69370 


1 12380 


17100 


68.4 


3-442 


2.192 


1.250 


7-854 


5574o 


42960 


67460 


118300 


18830 


77.1 


3.426 


2. 181 


1.245 


7-854 


5574o 


41780 


65640 


1 14980 


18230 


74-9 


3-554 


2-319 


1-235 


7-854 


59180 


435™ 


66690 


125220 


19690 


73-5 


3-569 


2.329 


1.240 


7-854 


59180 


43830 


67170 


126160 


19920 


74-i 


3.780 


2.520 


1.260 


7-854 


60480 


44580 


66870 


x 33730 


21450 


73-7 


3-765 


2.510 


1-255 


7-854 


60480 


44410 


66610 


133230 


21290 


73-4 



PLATE. 



3-559 


2.190 


1.369 


3.682 


54260 


40460 


65740 


105170 


39100 


74.6 


3-549 


2.184 


1.365 


3.682 


54260 


39420 


64060 


102490 


38000 


72.6 


3-829 


2.460 


1.369 


3.682 


54260 


39780 


61910 


111250 


41360 


73-3 


3-843 


2.471 


1-372 


3.682 


54260 


39060 


60740 


109400 


40770 


72.0 


3-843 


2.196 


1.647 


5-300 


54260 


37000 


64750 


86330 


26830 


68.2 


3-854 


2.202 


1.652 


5-3oo 


54260 


37050 


64840 


86430 


26940 


68.3 


4.118 


2.471 


1.647 


5-300 


54260 


37450 


62400 


93620 


29090 


69.0 


4.106 


2.464 


1.642 


5-300 


54260 


38040 


63390 


95130 


29470 


70.1 


4.656 


2.910 


1.746 


5-3oo 


59730 


41820 


66910 


111510 


36730 


70.0 


4.680 


2.925 


1-755 


5-300 


5973o 


42000 


67200 


1 1 2000 


37090 


70-3 


4.683 


3-031 


1.652 


5-3O0 


57870 


41040 


63410 


1 1 6340 


36260 


70.9 


4-938 


3-197 


1.741 


5-300 


59730 


40910 


63180 


1 16030 


38110 


68.5 


4-i5i 


2.214 


1-937 


7.216 


54260 


35000 


65620 


75000 


20130 


64-5 


4.114 


2.192 


1.922 


7.216 


54260 


34180 


64140 


73150 


19480 


63.0 


4.668 


2.626 


2.042 


7.216 


59730 


36870 


65540 


84280 


23850 


61 .7 


4.656 


2.619 


2.037 


7.216 


5973o 


38940 


69220 


89000 


25120 


65.2 


4.916 


2.895 


2.021 


7.216 


5973o 


38730 


65770 


94a 10 


26390 


64.8 


4.672 


2-745 


1.927 


7.216 


57870 


39010 


65660 


94580 


25260 


67.4 


5.076 


3.102 


1.974 


7.216 


5373o 


3796o 


62120 


97620 


26700 


70.6 


5-130 


3-!35 


i-995 


7.216 


58340 


39810 


65140 


102360 


28300 


68.2 


5-45o 


3-439 


2. on 


7.216 


5373o 


38920 


61670 


105470 


29390 


72.4 


5.290 


3-342 


1.948 


7.216 


56670 


39870 


63110 


108260 


29230 


70.4 



of riveting ; the two middle sections sheared in front *f rivets. 



6o6 



APPLIED MECHANICS. 



TABULATION OF SINGLE- 

t" STEEL 



f 


















No. of 
Test. 


Sheet Letters. 


Pitch. 


No. of 
Riv- 
ets. 


Width 

of 
Joint. 


Nominal 
Thickness. 


Size of 
Rivets 

and 
Holes. 


Actual 
Thick- 
ness of 
Plate. 


Lap. 












Plate. 


Covers. 








Plate. 


Covers. 
















in. 




in. 


in. 


in. 


in. 


in. 


in. 


1378 


D 


M 


M 


2* 


6 


15.00 


3/8 


1/4 


H*i 


•383 


2 


1379* 


D 


M 


M 


" 


6 


** 


" 


" 


" 




385 


2 


1380 


B 


J 


K 


2 


6 


12.00 


" 


" 


4i*i 




388 


2 


1381 


E 


K 


K 


" 


6 


" 


" 


11 






381 


2 


1382 


B 


K 


K 


2* 


6 


12.75 


" 


" 






388 


2 


1383+ 


E 


G 


K 


" 


6 


" 


" 


(i 






383 


2 


1384 


F 


H 


H 


2i 


6 


13-49 


44 


' " 






381 


2 


13S5 


E 


N 


N 


,k 


6 


" 


44 


ii 






380 


2 


1386$ 


H 


L 


L 


*t 


6 


14-25 


44 


11 






368 


2 


1387 


G 


L 


M 


" 


6 


" 


ti 


44 






365 


2 


1388 


D 


M 


M 


*i 


6 


15.00 


44 


44 






385 


2 


1389 


D 


M 


M 


" 


6 


" 


44 


i< 






386 


2 


1390 


C 


G 


G 


2| 


5 


13.12 


44 


44 






372 


2 


ng*§ 


C 


H 


L 


" 


5 


" 


(i 


it 






369 


2 


139 2 


C 


L 


L 


2f 


5 


13-75 


M 


tt 






374 


2 


1393 


C 


L 


N 


" 


5 


" 


it 


44 






372 


2 


i394§ 


D 


I 


M 


2| 


5 


M-39 


41 


11 






386 


2 


1395 


D 


M 


M 


" 


5 


1 


II 


" 






383 


2 

_. <! 



* Test discontinued soon after passing maximum load. 
+ Test discontinued at maximum load. 
$ Test discontinued after passing maximum load. 
§ Test discontinued before fracture was complete. 



TABULATION OF RIVETED JOINTS. 



607 



RIVETED BUTT-JOINTS— Continued. 
PLATE— Continued. 



Sectional Area 
of Plate. 


Bearing 
Surface 

of 
Rivets. 


Shear- 
ing 
Area of 
Rivets. 


Tensile 
Strength 
of Plate 

per 
Square 
Inch. 


Maxi 


mum Stress on Joint 
Square Inch. 


per 


Effi- 
ciency 

of 
Joint. 


Plate. 


Covers. 


Tension 
on Gross 

Section 
oi Plate. 


Tension 

on Net 

Section 

of Plate. 


Compres- 
sion on 
Bearing 
Surface 

of 
Rivets. 


Shear- 
ing of 
Rivets. 


| sq. in. 


sq. in. 


sq. in. 


sq. in. 


lbs. 


lbs. 


lbs. 


lbs. 


lbs. 


| 5 745 


3-734 


2.01 1 


7.216 


5373o 


40560 


62400 


115860 


32290 


75-5 


5 775 


3-754 


2.021 


7.216 


5373o 


40700 


62620 


116280 


32570 


75-7 


4656 


2.328 


2.328 


9-425 


5973° 


345oo 


69010 


69010 


17050 


57-8 


4-572 


2.286 


2.286 


9.425 


58340 


33440 


66880 


66880 


16220 


57-3 


4-947 


2.619 


2.328 


9-425 


5973o 


3559o 


67230 


75640 


18680 


59-6 


4.883 


2-585 


2.298 


9-425 


58340 


3473° 


65610 


73800 


17990 


59-5 


5-'40 


2.854 


2.286 


9-425 


54290 


3549o 


63930 


79810 


19360 


65-4 


5 126 


2.846 


2.280 


9-425 


58340 


35840 


64550 


80570 


19490 


61.4 


5.244 


3.036 


2.208 


9-425 


56670 


37010 


63930 


87910 


20590 


65.3 


5.205 


3-oi5 


2.190 


9-425 


53840 


36750 


63450 


87350 


20300 


68.2 


5-775 


3-465 


2.310 


9-425 


5373o 


3749o 


62480 


93710 


22970 


69.8 


5-79C 


3-474 


2.316 


9-425 


53730 


3736o 


62260 


9339o 


21890 


69-5 


4.881 


3.021 


1.860 


7-854 


57870 


39000 


63010 


102340 


24240 


67.4 


4.841' 


2.996 


1.845 


7-854 


57870 


39520 


63850 


103690 


24360 


68.3 


5- I 43 


3- 2 73 


1.870 


7-854 


57870 


39840 


62590 


109540 


26080 


68.9 


5.111 


3 2 5i 


1.860 


7.854 


57870 


40420 


63550 


1 11070 


26310 


69.8 


5-555 


3-625 


1.930 


7-854 


5373o 


3934o 


60290 


113240 


27830 


73-2 


5-496 


3.58i 


1.915 


7-854 


5373o 


40300 


61850 


115660 


28200 


75 



608 APPLIED MECHANICS. 



SINGLE-RIVETED BUTT-JOINTS, STEEL PLATE. 
DESCRIPTION OF TESTS AND DISCUSSION OF RESULTS. 

" The following tests complete a series of two hundred and 
sixteen single-riveted butt-joints in steel plates, in which the 
thickness of the plates ranged from J" to J ", and the size of 
the rivets from T 9 -g-" to \-^" diameter. 

The plates were annealed after shearing to size, the edges 
opposite the joint milled to the finished width ; the holes were 
drilled and rivets machine-driven. Iron rivets were used 
throughout, except in some of the f" joints. 

Tensile tests of the plates and rivet-metal, together with 
the tests of the joints in J" and § " plate, are contained in the 
Report of Tests of 1885, Senate Document No. 36, Forty-ninth 
Congress, first session. 

The tests herewith presented comprise the details and tab- 
ulation of joints in \ h ', §", and J" thickness of plate, a portion 
of which were tested hot. 

The gauged length in which elongations and sets were 
measured was $"; 2\" each side of the centre line of the joint. 

During the progress of testing the same characteristics were 
displayed which were referred to in the previous report. The 
joints were very rigid under the early loads. This rigidity is 
overcome by loads which exceed the friction between the plate 
and covers, after which the stretching proceeded slowly with 
some fluctuations till elongation of the metal of the net section 
became general ; the metal under compression in front of the 
rivets yielding, also the rivets themselves. 

The behavior of joints in different thicknesses of plate is 
substantially the same, and an examination of the results shows 
that when exposed to similar conditions the strength per unit 



SINGLE-RIVETED BUTT-JOINTS, STEEL PLATE. 609 

of fractured metal is nearly the same, whether J" or J" plate is 
used. 

It will not be understood from this, however, that as a con- 
sequence the same efficiency may be obtained in different 
thicknesses of plate for single-riveted work, because it will be 
seen that certain essential conditions change as we approach 
the stronger joints in different thicknesses of plate. 

A riveted joint of the maximum efficiency should fracture 
the plate along the line of riveting, for it is clear that if failure 
occurs in any other manner, as by shearing the rivets or tear- 
ing out the plate in front of the rivet-holes, there remains an 
excess of strength along the line of riveting, or in other words 
along the net section of metal — if in a single-riveted joint-- 
which has not been made use of ; but when fracture occurs 
along the net section an excess of strength in other dircctiom- 
is immaterial. 

If the strength per unit of metal of the net section was con- 
stant, it would be a very simple matter to compute the effi- 
ciency of any joint, as it would merely be the ratio of the net 
to the gros9 areas of the plates. 

The tenacity of the net section, however, varies, and this 
variation extends over wide limits. 

In the present series there is an excess in strength of the 
net section over the strength of the tensile test-pieces in all 
joints. 

Special tables have been prepared showing this behavior. 

The efficiencies shown in Table No. 1 are obtained by divid- 
ing the tensile stress on the gross area of plate by the tensile 
strength of the plate as represented by the strength of the ten- 
sile test-strip, stating the values in per cent of the latter. 

Table No. 2 exhibits the differences between the efficien- 
cies of the joints and the ratios of net to gross areas of plate. 
If the tenacity 'of net section remained constant per unit of 



6kO 



APPLIED MECHANICS. 



area, the efficiencies in Table No. i would, as above explained, 
be identical with the ratios of net to gross areas of plate, and 
the values in this table reduced to zero. 

Table No. 3 shows the excess in strength of the net section 
of the joint over the strength of the tensile test-strip in per 
cent of the latter. 

Table No. 4 exhibits the compression on the bearing-surface 
of the rivets in connection with the excess in tensile strength 
of the net section of plate. 

Table No. 1 is valuable in showing at once the value of 
different joints wherein the pitch of the rivets and their diame- 
ters vary. 

It is seen there is considerable latitude allowed in the choice 
of rivets and pitch without materially changing the efficiency 
of the joint ; thus in J" plate, 

f" rivets (driven), i-J" pitch, 72.4 per cent efficiency, 
f" rivets (driven), 23-" pitch, 73.3 per cent efficiency, 
\ ft rivets (driven), 2§" pitch, 71.5 per cent efficiency, 
1" rivets (driven), 2-j-" pitch, 70.3 per cent efficiency, 
1" rivets (driven), 2 J" pitch, 73.8 per cent efficiency, 

give nearly the same results. 

In these examples the ratios of net to gross areas of plate 
range from 60 to 67 per cent, while the rivet-areas range from 
.3067 square inch to .7854 square inch. The actual areas of 
net sections of plate and rivets are as follows : 





§" rivets. 


£" rivets. 


I" rivets. 


1" rivets. 


Rivets 


sq. in. 
,^067 


sq. in. 
.4418 

2.207 


sq. in. 
.6013 

2.232 


sq. in. 
.7854 
j 2.259 
( 2.319 


Plate 


I.486 



SINGLE-RIVETED BUTT-JOINTS, STEEL PLATE. 6 1 I 



The areas of the rivets stand to each other as the following 
numbers : 



ioo 



144 



96 



and the net areas of the plate to each other as 

149 150 



100 



256 



152 
156 



From these illustrations it appears that to attain the same 
degree of efficiency in this quality of metal, although that 
efficiency is probably not the highest attainable, a fixed ratio 
between rivet metal and net section of plate is not essential. 

In \" plate with -J" rivets the efficiencies of the joints tested 
cold are nearly constant over the range of pitches tested. 

The efficiencies and the ratio of net to gross areas of plate 
are as follows : 





Pitch. 


i£" 


1" 


2$" 


2\" 


Efficiency. 

Ratio of areas . . 


per cent. 

64-5 
53-4 


per cent. 
66.3 
56.3 


per cent. 
66.3 
58.9 


per cent. 
66.4 
6l. I 



In this we have illustrated a case which, in passing from 
the widest pitch, having 61.1 per cent of the solid plate left, to 
the narrowest pitch, which had 53.4 per cent of the solid plate, 
the gain or excess in strength in the net section almost exactly 
compensated for the loss of metal. 

In Table No. 3 the average of all the joints shows the high- 
est per cent of excess of strength in the narrowest pitch, and a 
tendency to lose this excess as the pitch increases. 

Tests of detached grooved specimens show the same kind 



612 



APPLIED MECHANICS. 



of behavior, but as they are not subject to all the conditions 
found in a joint, the analogy does not extend very far. 

The maximum gain in strength on the net section, not for 
the time being regarding the hot joints, and disregarding the 
exceptionally high value of joint No. 1339, J" plate, was 21.2 
per cent, the minimum value 2.5 per cent of the tensile test- 
strip. In other forms of joints, and with punched holes in both 
iron and steel plate, illustrations are numerous in which there 
have been large deficiencies, the metal of the net section fall- 
ing far below the strength of the plate. 

It is believed to have been amply shown that increasing 
the net width diminishes the -apparent tenacity of the plate, 
although other influences may tend to counteract this tendency 
in some joints. 

In order to compare the excess in strength of one thickness 
of plate with another having the same net widths, we have the 
following table, rejecting those joints that failed otherwise 
than along the line of riveting in making these averages: — 



Thickness of Plate. 


Width of plate between rivet-holes. 


i" 


i£" 


ij" 


if" 


i|" 


if" 


if" 


if" 


2" 


1" 

£." 


P. ct. 

16.7 
18.4 


P. ct. 

12.6 

13-7 
14-3 
163 
151 


P. ct. 

11. 4 
12.7 

9-3 
14.2 
13-8 


P. ct. 

12.0 
i3-5 
10.7 
i4-5 
14. 1 


P. Ct. 

13-4 
14.6 
q.i 

14.6 
7.6 


P. Ct. 

8.9 
12.9 

8.8 
12.7 

IT. 8 


P. ct. 

"•5 
9.0 
8.2 

9-9 

10. 


P. Ct. 

• 13-1 
13-6 
12.2 

9.8 
10. 1 


P. ct. 

10.6 

3-5 


£" 


16.7 


£" 




i" 

Average of all thick- 
nesses 


11. 4 


16.2 


14.4 


12.3 


12. g 


11. 9 


11. 


9-7 


11. 8 


7.0 



The excess in strength is generally well maintained in each 
of the several thicknesses, and were it possible to retain the 
same ratio of net to gross areas of plate, and at the same time 



SINGLE-RIVETED BUTT-JOINTS, STEEL PLATE. 613 

equal net widths between rivets, it would seem from this point 
of view feasible to obtain the same degree of efficiency in thick 
as in thin plates. 

The following causes, however, tend to prevent such a con- 
summation. 

For equal net widths thick plates require larger rivets to 
avoid shearing than thin ones, the diameters of the rivets being 
somewhat increased for this cause, and again because it has 
become necessary to increase the metal of the net section in 
order to retain a suitable ratio of net to gross areas of plate. 

There results from these considerations such an increase in 
net width of plate that the excess in strength displaj^ed by 
narrower sections is lost, and consequently the result is a joint 
of lower efficiency. 

The data relating to the influence of compression on the 
bearing-surface of the rivets, on the tensile strength of the 
plate, as shown by Table No. 4 are more or less conflicting. 
However, in the \" plate, in which the most intense pressures 
are found, there is seen a pronounced increase in tensile strength 
as the pressures diminish in intensity. 

It is probable that the effects of intense compression would 
be more conspicuous in a less ductile metal, or one in which 
the ductility had been impaired by punched holes or otherwise. 

A number of joints were tested at temperatures ranging 
between 200 and 700 ° Fahr. 

The heating was done after the joints were in position for 
testing, by means of Bunsen burners, arranged in a row par- 
allel to and under the line of riveting. 

The temperature was determined with a mercurial ther- 
mometer, the bulb of which was immersed in a bath of oil, 
contained in a pocket drilled in the middle rivet of the joint. 

When at the required temperature the thermometer was 
removed from the joint, a dowel was driven into the pocket to 



6 14 APPLIED MECHANICS. 

compensate for the metal of the rivet which had been removed 
by the drill, and then loads applied and gradually increased up 
to the time of rupture. 

Three joints, Nos. 1423, 1426, and 1430, were tested with- 
out dowels in the oil-pockets. 

The method of heating was to raise the temperature of the 
joint, as shown by the thermometer, a few degrees above the 
temperature at which the test was made, shut off the gas- 
burners, and allow the temperature to fall to the required limit. 
The temperature fell slowly, draughts of cold air being excluded 
from the under side of the joint by the hood which covered 
the gas-burners ; the upper side and edges of the joint were 
covered with fine dry coal-ashes. 

The results show an increase in tensile strength when heated 
over the duplicate cold joints at each temperature except 200 
Fahr. 

From 200 there was a gain in strength up to 300 , when 
the resistance fell off some at 350 , increased again at 400 , and 
reached the maximum effect observed at 500 Fahr. ; from this 
point the strength fell rapidly at 6oo° and 700 . 

In per cent of the cold joint there was a loss at 200 of 3.2 
per cent, the average of three joints ; at 500 the gain was 22.6 
per cent, the average of four joints. The maximum and mini- 
mum joints at this temperature showed gains of 27.6 per cent, 
and 18.3 per cent, respectively. 

The highest tensile strength on the net section of plate was 
found in joint No. 1433, tested at 500 Fahr., where 81050 
pounds per square inch was reached against a strength of 58000 
pounds per square inch in the cold tensile test-strip. 

The hot joints showed less ductility than the cold ones, 
those tested at 200 Fahr. not being exempt from this behav- 
ior, although there was no near approach to brittleness in any. 

Three joints, Nos. 141 8, 1420, and 1424, were heated ; 



SINGLE-RIVETED BUTT-JOINTS, STEEL PLATE. 615 

strained when hot with loads exceeding the ultimate strength 
of their- duplicate cold joints; the loads were released, and 
after having cooled to the temperature of the testing-room 
(No. 1424 cooled to 150 Fahr.) were tested to rupture, and 
were found to have retained substantially the strength due 
their temperature when hot. 

In order to ascertain that the time intervening between hot 
straining and final rupture did not contribute towards the ele- 
vation in strength, joint No. 1434 was strained in a similar 
manner with a load approaching rupture, after which a period 
of rest was allowed and then ruptured without material gain 
in strength. 

A peculiarity of the joints fractured at 400 and higher tem- 
peratures was the comparatively smooth surface of the frac- 
tured sections, and which took place in planes making angles 
of about 50 with the rolled surface of the plate. 

The shearing-strength of the iron rivets was also increased 
by an elevation of temperature. 

The rivets in joint No. 1410 at the temperature of 350 
sheared at 43060. pounds per square inch, while in the dupli- 
cate cold joint No. 141 1 they sheared at 38530 pounds per 
square inch, and the rivets in joint No. 1398 at 300 Fahr. 
were loaded with 46820 pounds per square inch and did not 
shear. 

Other examples, where some of the rivets sheared and the 
plate fractured in part, showed corresponding gains in shearing- 
strength. 

The almost entire absence of granular fractures in these 
tests is a feature too important to pass by without special men 
fcion." 



6i6 



APPLIED MECHANICS. 



TABULATION OF SINGLE- 

STEEL PLATE. 



No. 

of 

Test. 


Sheet Letters. 


Pitch. 


No. 

of 

Rivets. 


Width 
of 


Nominal Thick- 
ness. 


Size of 
Rivets 

and 
Holes. 


Actual 
Thick- 
ness of 
Plate. 


Lap. 


Plate. 


Covers. 


Joint. 


Plate. 


Covers. 










in. 




in. 


in. 


in. 


in. 


in. 


in. 


1396 


R 


R 


R 


if 


6 


10.50 


1/2 


5/16 


tt** 


.481 


2 


*397 


R 


R 


R 


" 


6 


' 




" 


44 


44 




.484 


2 


1398 


R 


R 


R 


«* 


6 


11 


25 


" 


44 


44 




484 


2 


*399 


R 


R 


R 


" 


6 


' 




" 


" 


44 




483 


2 


1400 


R 


R 


R 


2 


6 


12 


00 


" 


. 4 ' 


44 




.486 


2 


1 401 


R 


S 


T 


u 


6 


4 




" 


" 


44 




483 


2 


1402 


R 


R 


R 


1* 


6 


11 


25 


it 


«t 


tt&f 




481 


2 


1403 


R 


R 


R 


" 


6 


' 




" 


44 


44 




486 


2 


1404 


R 


R 


R 


2 


6 


12 


00 


" 


" 


44 




486 


2 


1405 


R 


S 


S 


" 


6 


4 




(C 


44 


44 




487 


2 


1406 


S 


s 


S 


a* 


6 


12 


75 


»« 


44 


Itt 




470 


2 


1407 


S 


s 


S 


" 


6 


• 




" 


44 


tt 




47i 


2 


1408 


T 


Q 


Q 


2i 


6 


13 


bo 


(i 


t« 


11 




486 


2 


1409 


T 


Q 


Q 


" 


6 


1 




" 


44 


44 




482 


2 


1410 


T 








2f 


6 


14 


25 


" 


ti 


44 




481 


2 


1411 


T 








" 


6 


4 




" 


44 


44 




485 


2 


1412 


R 


R 


R 


2 


6 


12 


00 


" 


44 


tt*i 




484 


2 


I4'3 


R 


R 


R 


" 


6 


4 




" 


44 


44 




481 


2 


1414 


S 


s 


S 


2 fr 


6 


12 


75 


" 


44 


44 




472 


2 


*4>5 


S 


s 


S 


" 


6 


' 




" 


tt 


44 




468 


2 


1416 


S 





Q 


8l 


6 


13 


5o 


it 


it 


44 




468 


2 


1417 


T 


Q 


Q 


" 


6 


' 




" 


44 


44 




482 


2 


1418 


T 








2| 


6 


H 


25 


" 


it 


44 




481 


2 


1419 


T 





O 


44 


6 


1 




" 


14 


44 




482 


2 


1420 


U 


p 


P 


2* 


6 


IS 


00 


" 


44 


it 




479 


2 


1421 


U 


p 


P 


" 


6 


4 




" 


44 


tt 




483 


2 


1422 


s 


s 


s 


2* 


5 


13 


13 


" 


tt 


tt 




469 


2 


1423 


s 


s 


s 


" 


5 


' 




it 


44 


44 




473 


2 


1424 


T 








a* 


5 


13 


75 


" 


44 


it 




484 


2 


1425 


T 








" 


5 


' 




" 


44 


" 




483 


2 


1426 


s 


' s 


s 


a* 


6 


12 


75 


it 


44 


ife&iB 




474 


2 


1427 


R 


s 


s 


" 


6 


' 




" 


44 


. " 




475 


2 


1428 


T 


Q 


Q 


ai 


6 


13 


50 


" 


" 


<t 




479 


2 


1429 


S 





Q 


" 


6 


' 




*' 


«t 


44 




465 


2 


1430 


U 








8f 


6 


14 


25 


" 


44 


44 




484 


2 


M3 1 


u 








" 


6 


4 




" 


44 


44 




483 


2 



TABULATION OF RIVETED JOINTS. 



6iy 



RIVETED BUTT-JOINTS. 



STEEL PLATE. 



Sectional Area 
of Plate. 



Gross. 



sq in. 

5.082 
5-445 
5-439 
5.842 
5-79 6 
5-4i6 
5-477 
5-832 
5-854 
5-997 
6.010 
6.561 
6.512 
6.859 
6.916 
5-8i3 
5-772 
6.023 

5-9 6 7 
6.327 
6.512 
6.859 
6.883 
7-194 
7-245 
6.167 
6 220 
6.660 
6 632 
6.053 
6.042 
6.471 
6.278 
6.897 
6.852 



Net. 



(Bearing 

! Surface 

of 

Rivets. 



sq. in. 
2.886 
2.904 
3.267 
3.266 
3-655 
3 . 622 
2.891 
2.926 
3.281 
3-297 
3-529 
3-537 
4.010 
3.982 
4-334 
4-37° 
2.909 
2.886 
3-i9i 
3-159 
3-5 T 9 
3.620 

3-973 
3-991 
4.320 

4-347 
3 822 
3-855 
4.240 
4.217 
2.853 
2.836 
3.238 
3-139 
3.620 
3.632 



sq. in. 
2.165 
2.178 
2.178 
2.173 
2.187 
2.174 
2-525 
2551 
2-55' 
2-557 
2.468 
2.473 
2.551 
2-53° 
2-525 
2.546 
2.904 
2.886 
2.832 
2.808 
2.808 
2.892 



2.874 
2.898 
2-345 
2-365 
2.420 
2-415 
3.200 
3.206 
3-233 
3-139 
3-277 
3.260 



Shear- 
ing 
Area of 
Rivets. 



sq. in. 
5-3oi 
5-3oi 
5-3oi 
5 -301 
5-3oi 
5- 301 
7.216 
7.216 
7.216 
7.216 
7.216 
7.216 
7.216 
7.216 
7.216 
7.216 
9-425 
9-425 
9425 
9-425 
9-425 
9-425 
9.425 
9-425 
9-425 
9-425 
7-854 
7-854 
7-854 
7-854 
11.928 
11.928 
11.928 
11.928 
n.928 



Tensile 


Max. Stress on Joint per Sq. In. 


C c 

y ._ 
"3,° 

W 


ha 


Strength 
of Plate 

per 
Sq. In. 


Tension 

on Gross 

Section of 

Plate. 


Tension 
on Net 
Sectionof 
| Plate. 


Comp. on 
Bearing 
Surface 

of Rivets. 


<U b/3> 


lbs. 


lbs. 


lbs. 


lbs. 


lbs. 






57180 


3775o 


G6070 


88080 


3597o 


66.1 


200 


.57180 


38980 


68220 


90960 


37370 


68.1 




57180 


45560 


75960 


1 13960 


46820 


79.6 


300 


57180 


39260 


65400 


98290 


40290 


68.6 




57!8o 


37000 


59140 


98830 


40770 


64.7 




57180 


39420 


63080 


105100 


43100 


68.9 


350 


57i8o 


36890 


69110 


79130 


27690 


64-5 




57180 


382*50 


71600 


81730 


29030 


66.9 


250 


57180 


379io 


67380 


86670 


30640 


66.3 




57180 


4373o 


7 7650 


100120 


3548o 


76.4 


300 


59050 


4479o 


76110 


108830 


37220 


76.0 


400 


59050 


39210 


66630 


953io 


32660 


66.3 




60000 


39850 


65210 


102500 


36230 


66.4 




60000 


46610 


76220 


119980 


42060 


77-6 


500 


60000 


453oo 


71690 


123050 


43060 


75-5 


350 


60000 


40050 


63390 


108800 


38530 


66.7 




57180 


35920 


71770 


71900 


22150 


62.8 


250 


57180 


3439° 


68780 


68780 


21060 


60.1 




59050 


35000 


66020 


74430 


22360 


59-2 




59050 


40250 


76030 


85540 


25480 


68.1 


300 


59050 


34660 


62320 


78090 


23160 


60.3 


200 


60000 


36950 


66480 


83220 


25530 


61.5 


V 


60000 


437io 


75460 


103880 


31810 


72.8 


(*) 


60000 


38720 


66770 


92150 


28270 


64-5 




58000 


44840 


74670 


112250 


34230 


77-3 


(t) 


58000 


38740 


64570 


96850 


297P0 


66.9 




59050 


3973o 


64110 


104490 


31200 


67.2 




59050 


45420 


73250 


119450 


35840 


76.9 


400 


60000 


48950 


76890 


137110 


41510 


81.5 


(*) 


60000 


40600 


63860 


1 1 1490 


34280 


67.6 




59050 


35070 


74410 


66340 


18630 


59-3 


300 


59050 


30420 


64810 


5733° 


15410 


5i-5 




60000 


40330 


80620 


80730 


21880 


67.2 


350 


59050 


33420 


66840 


66840 


17590 


56.5 




58000 


36390 


65150 


76590 


21040 


62.7 


700 


58000 


33660 


63870 


71160 


19450 


58.0 





* Strained while at temperature of 400 Fahr., and allowed to cool before rupture. 
+ Strained while at temperature of 500 Fahr., and allowed to cool before rupture. 
% Strained while at temperature of 500 Fahr., then cooled to 150 Fahr., and ruptured. 



6i8 



APPLIED MECHANICS. 



TABULATION OF SINGLE- 
STEEL PLATE— Continued. 







Sheet 










Nominal 










Letters 










Thickness. 








No. 

of 

Test. 








Pitch. 


No. 
of 

Riv- 
ets. 


Width 

of 
Joint. 




Size of 
Rivets 

and 
Holes. 


Actual 
Thick- 
ness of 
Plate. 


Lap. 














Plate. 


Covers. 








Plate. 


Covers. 
















in. 




in. 


in. 


in. 


in. 


in. 


in. 


1432 


U 


P 


p 


*i 


6 


15.00 


1/2 


5/i6 


IT3& i§ 


.484 


2 


1433 


U 


P 


p 


" 


6 




1 




" 


" 




481 


2 


1434 


R 


S 


Q 


A 


5 , 


13 


13 







44 




472 


2 


1435 


S 


s 


s 


" 


5 




' 




14 


44 




475 


2 


1436 


T 








*1 


5 


13 


75 




it 


44 




482 


2 


1437 


T 








" 


5 




4 




" 


tt 




482 


2 


1438 


U 


P 


p 


A 


5 


14 


38 




" 


44 




484 


2 


1439 


U 


p 


p 


it 


5 




' 




44 


44 




485 


2 


1440 


U 


p 


p 


3 


5 


15 


00 




44 


41 




482 


2 


1441 


u 


p 


p 


" 


5 




1 




41 


41 




483 


2 


1442 


u 


p 


p 


si 


5 


15 


63 




(1 


41 




483 


2 


1443 


u 


p 


p 


" 


5 




' 




44 


44 




484 


2 


1444 


V 


E 


E 


If 


6 


11 


25 


5/8 


3/8 


«*s 




621 


2 


1445 


V 


B 


E 


44 


6 




' 




it 


44 




624 


2 


1446 


V 


E 


E 


2 


6 


12 


00 




44 


tt 




616 


2 


1447 


V 


E 


E 


" 


6 




4 




44 


II 




624 


2 


1448 


V 


E 


E 


*i 


6 


12 


75 




44 


44 




621 


2 


1449 


V 


E 


E 


(( 


6 




1 




44 


44 




624 


2 


1450 


w 


F 


G 


A 


6 


13 


50 




44 


44 




610 


2 


1451 


w 


G 


G 


" 


6 




1 




44 


44 




611 


2 


i45 2 


V 


E 


E 


2 


6 


12 


00 




it 


H*i 




624 


2 


1453 


V 


E 


E 


" 


6 




4 




" 


" 




620 


2 


1454 


V 


E 


E 


*! 


6 


12 


75 




44 


44 




622 


2 


1455 


V 


E 


E 


" 


6 ! 




t 




tt 


41 




618 


2 


i45 6 


w 


G 


F 


*i 


6 


13 


50 




44 


44 




612 


2 


1457 


w 


G 


G 


" 


6 




' 




tt 


44 




611 


2 


1458 


w 


I 


1 


*i 


6 


14 


25 




1 


44 




610 


2 


1459 


w 


H 


H 


" 


6 




' 




44 


44 




608 


2 


1460 


X 


I 


I 


^ 


6 


15 


00 




u 


44 




617 


2 


1461 


X 


J 


I 


" 


6 




' 




44 


44 




618 


2 


1462 


V 


F 


F 


A 


5 


13 


13 




it 


41 




630 


2 


1463 


w 


F 


F 


" 


5 




' 




44 


44 




608 


2 


1464 


V 


E 


E 


*S 


6 


12 


75 




44 


lT«&l| 




624 


2 


1465 


V 


D 


E 


" 


6 




' 




11 


44 




623 


2 


1466 


w 


G 


G 


I? 


6 


13 


•50 




44 


44 




613 


2 


1467 


w 


n T y 


G 




6 






ii 






606 


2 



TABULATION OF RIVETED JOINTS. 



619 



RIVETED BUTT-JOINTS— Continued. 

STEEL PLATE— Continued. 



Sectional Area 
of Plate. 


Bearing 
Surface 

of 
Rivets. 


Shear- 
ing 
Area of 
Rivets. 


Tensile Strength of 
Plate per Square 
Inch. 


Maximum Stress on Joint per 
Square Inch. 


c 
'5 


>. 



a 

.a 


a a 




Tension on 
Gross Sec- 
tion of 
Plate. 


Tension on 
Net Sec- 
tion of 
Plate. 


Compression 
on Bearing 
Surface of 
Rivets. 


"o 

OX 9J 

>5 > 




Gross. 


Net. 


sq. in. 


sq. in. 


sq. in. 


sq. in. 


lbs. 


lbs. 


lbs. 


lbs. 


lbs. 






7.270 


4-043 


3.227 


11.928 


58000 


36140 


65000 


81430 


22030 


62.3 




7-215 


3-968 


3-247 


n.928 


58000 


4449° 


81050 


99040 


26960 


76.7 


500 


6.193 


3-538 


2-655 


9.940 


59050 


36670 


64190 


85540 


22850 


62.0 




6.232 


3-56o 


2.672 


9.940 


59050 


36200 


63370 


84420 


22690 


61.2 




6.632 


3.921 


2. 711 


9.940 


60000 


42430 


71610 


103580 


28250 


70.5 


600 


6.632 


3-92i 


2. 711 


9.940 


60000 


38720 


65440 


9473o 


25840 


645 




6.965 


4-243 


2.722 


9.940 


58000 


46630 


76550 


1 19320 


32680 


80.3 


500 


6.974 


4.246 


2.728 


9.940 


58000 


38900 


63890 


99400 


27290 


67.0 




7.230 


4-519 


2. 711 


9.940 


58000 


39180 


62690 


104500 


28500 


67.5 


200 


7.250 


4.528 


2.722 


9.940 


58000 


40360 


65060 


108220 


29630 


70.0 




7 564 


4.837 


2.717 


9.940 


574io 


38570 


60450 


107610 


29410 


67.2 




7-565 


4-843 


2.722 


9.940 


574io 


3^160 


61 1 70 


108830 


29800 


68.2 




6.986 


3.726 


3.260 


7.216 


55000 


3375° 


63280 


72330 


32670 


60.1 




7.020 


3-744 


3.276 


7.216 


55000 


3453° 


64740 


74000 


3359o 


62.7 




7-393 


4.158 


3-234 


7.216 


55000 


36760 


65340 


84010 


37650 


66.0 




7.488 


4.212 


3.276 


7.216 


55000 


35120 


62440 


80280 


36440 


63.8 




7.9x8 


4-658 


3.260 


7.216 


55000 


4i93o 


71270 


101840 


46010 


76.2 


300 


7956 


4.680 


3.276 


7^216 


55000 


36800 


62560 


89370 


40570 


66.9 




8.241 


5-039 


3.202 


7.216 


57290 


39320 


64290 


101180 


44900 


68.6 


400 


8.249 


5.042 


3.207 


7.216 


57290 


36850 


60290 


94790 


42130 


64-3 


600 


7.488 


3-744 


3-744 


9-425 


55000 


32080 


64150 


64150 


25480 


58.3 




7.440 


3.720 


3.720 


9-425 


55000 


32060 


64110 


64110 


25300 


58.3 






7-93 1 


4.199 


3-732 


9-425 


55000 


34120 


64440 


72510 


28710 


60.0 






7.880 


4.172 


3.708 


9-425 


55000 


34000 


64220 


72250 


28420 


61.8 






8.262 


4-59o 


3.672 


9-425 


57290 


36490 


65680 


82110 


32000 


63.6 






8.249 


4-583 


3.666 


9-425 


57290 


36020 


64830 


81040 


31520 


62.8 






8.662 


5.002 


3.660 


9-425 


57290 


37720 


65310 


89260 


38490 


65.8 






8.664 


5.016 


3-648 


9-425 


57290 


37540 


64850 


89170 


345'o 


65.5 






9- 2 55 


5-553 


3.702 


9-425 


5594o 


37300 


62160 


93250 


36630 


66.6 






9.282 


5-574 


3.708 


9-425 


5594o 


37000 


61610 


92620 


36440 


66.1 






8.259 


5.109 


3150 


7-854 


55000 


3578o 


57840 


93810 


37620 


65.0 






7-965 


4-925 


3 040 


7-854 


57290 


36960 


5977o 


96840 


37500 


64.5 






7-95° 


3-738 


4.212 


11.928 


55000 


31090 


66130 


58690 


20720 


56.5 






7-949 


3-744 


4.205 


11.928 


55000 


31090 


66020 


58780 


20720 


56.5 






8.269 


4-i3i 


4-I38 


11.928 


57290 


33150 


66350 


66240 


22980 


57-8 






8.t8i 


4.090 


4.091 


11.928 


5594o 


33240 


66250 


66240 


22720 


58.0 







620 



APPLIED MECHANICS. 



TABULATION OF SINGLE^ 

STEEL PLATE— Continued. 



No. 

of 

Test. 


Sheet Letters. 


Pitch. 


No. 

of 

Rivets. 


Width 

of 
Joint. 


Nominal 
Thickness. 


Size of 
Rivets 

and 
Holes. 


Actual 
Thick- 
ness of 
Plate. 


Lap. 


Plate 


Covers. 


Plate. 


'Covers. 










in. 




in. 


in. 


in. 


in. 


in. 


in. 


1468 


W 


H 


I 


*t 


6 


14.25 


5/8 


3/8 


ire & is 


.613 


2 


1469 


W 


I 


Nft" 


44 


6 


" 


" 


" 


44 


.609 


2 


1470 


X 


J 


I 


2i 


6 


15.00 


" 


44 


44 


.619 


2 


147 1 


X 


J 


I 


" 


6 


" 


" 


44 


44 


.616 


2 


I 47 2 


V 


K 




2f 


5 


i3-*3 


tt 


44 


tt 


.628 


2 


M73 


w 


F 


F 


« 


5 


" 


" 


tt 


" 


.609 


2 


1474 


w 


H 


D 


2f 


5 


13-75 


" 


tt 


tt 


.609 


2 


1475 


w 


G 




" 


5 


" 


44 


44 


44 


.610 


2 


I476 


w 


I 


I 


2j 


5 


14.38 


44 


44 


44 


.610 


2 


1477 


w 


I 


I 


" 


5 


" 


" 


44 


it 


.6oy 


2 


I 47 3 


X 


I 


J 


3 


5 


15.00 


" 


44 


it 


.616 


2 


1479 


X 


I 


J 


i< 


5 


" 


" 


44 


tt 


.623 


2 


T480 


G 


E 


E 


3t 


4 


12.50 


" 


tt 


44 


.625 


2 


I481 


H 


E 


E 


'» 


4 


" 


44 


(( 


44 


.621 


2 


I482 


Z 


K 


K 


2 


6 


12.00 


3/4 


7/16 


l§&x 


•736 


2 


1483 


Z 


K 


K 


" 


6 


" 


" 


44 


tt 


•757 


2 


| 1484 


Z 


K 


K 


2* 


6 


12.75 


" 


44 


tt 


.742 


2 


1485 


Z 


P 


P 


" 


6 


tt 


ti 


tt 


tt 


.762 


2 


i486 


Z 


L 


M 


2* 


6 


13-50 


" 


44 


ft 


•749 


2 


I487 


Z 


L 


M 


" 


6 


" 


" 


(t 


tt 


.764 


2 


14 8 


Z 


N 


N 


2| 


6 


14-25 


it 


44 


tt 


•745 


2 


1489 


Z 


O 


N 


" 


6 


l( 


" 


44 


tt 


•735 


2 


l 4 yo 


z 


K 




2l 


6 


12.75 


" 


44 


I& * l| 


•723 


2 


1491 


z 


P 


P 


" 


6 


" 


ii 


44 




752 


2 


1492 


z 


M 


L 


zi 


6 


13-50 


" 


44 


44 


•736 


2 


1493 


z 


M 


M 


" 


6 


" 


" 


tl 


tt 


•754 


2 


1494 


z 


N 


O 


2f 


6 


14.25 


44 


44 


44 


.760 


2 


1495 


z 


O 


O 


" 


6 


" 


14 


It 


44 


.760 


2 


1496 


Y 


Q 


P 


«* 


6 


15.00 


44 


44 


ft 


•745 


2 


M97 


Y 


P 


P 


44 


6 


(« 


44 


tt 


tt 


•725 


2 


1498 


Z 


L 


L 


2| 


5 


13-13 


44 


" 


ft 


•733 


2 


1499 


Z 


L 


L 


" 


5 


" 


44 


If 


44 


•744 


2 


1500 


Z 


N 


M 


2} 


5 


13-75 


44 


" 


" 


.762 


2 


1 501 


Z 


N 


N 


" 


5 


" 


44 


" 


" 


.727 


2 


1502 


Z 


O 


P 


2| 


5 


14.38 


tt 


tt 


tt 


.722 


2 


1503 


Z 


O 


O 




5 








tt 


.741 


2 



TABULATION OF RIVETED JOINTS. 



62 1 



RIVETED BUTT-JOINTS— Continued. 
STEEL PLATE— Continued. 



1 Sectional Area 
of Plate. 


Bearing 
Surface 

of 
Rivets. 


Shear- 
ing 
Area of 
Rivets. 


Tensile 
Str'gth 

of Plate 

per 
Sq. In. 


Max. Stress on Joint per 


Sq. In. 


<4-l 
O 

v 

yi— 1 

&3 . 


4) 

ti w rt 
£ cfc. 

§£* 

,"oQ 
H 


Tension 
on Gross 
Section 
of Plate. 


Tension 
on Net 
Section 
of Plate. 


Comp. on 
Bearing 
Surface 

of Rivets. 


be 
c i2 


: Gross. 


Net. 


sq. in. 


sq. in. 


sq. in. 


sq. in. 


lbs. 


lbs. 


lbs. 


lbs. 


lbs. 






8-735 


4-597 


4.138 


11.928 


572>o 


34260 


65110 


72330 


25090 


59-8 




8.690 


4-579 


4. in 


11.928 


44 


3479o 


66030 


7354o 


25350 


60.7 




9.285 


5-107 


4.178 


11.928 


5594o 


34980 


63600 


77740 


27230 


62.5 




9.240 


5.082 


4-158 


11.928 


" 


34770 


63220 


77270 


26930 


62.1 




8.239 


4.706 


3-533 


9.940 


55000 


36350 


63640 


84770 


30130 


66.1 




7.978 


4-552 


3.426 


9.940 


57290 


37100 


65020 


86400 


29780 


64.7 




8.362 


4-936 


3.426 


9.940 


" 


38150 


61630 


93"o 


32090 


66.5 




8381 


4-950 


3-431 


9.940 


" 


38120 


61540 


93120 


32140 


66.5 




8.833 


5.402 


3-431 


9-94o 


" 


38620 


63110 


99410 


343io 


67.4 




8-739 


5-313 


3.426 


9.940 


" 


38180 


62830 


97430 


33580 


66.6 




9.240 


5-775 


3-465 


9.940 


5594o 


38480 


61570 


102630 


35770 


68.7 




9-345 


5.841 


3504 


9.940 


" 


38410 


61130 


102440 


36110 


68.6 




7-813 


5.000 


2.813 


7-952 


55000 


3734o 


58360 


103730 


36690 


67.9 




7-763 


4.968 


2-795 


7-952 


" 


38440 


60060 


106760 


37520 


69.9 




8.847 


4-43 1 


4.416 


9-425 


59000 


31990 


63870 


64090 


30030 


54-2 




9.099 


4-557 


4-542 


9-425 


it 


31980 


63860 


64070 


30870 


54-2 




9-475 


5.023 


4-452 


9.425 


44 


34440 


64960 


73920 


34620 


58.3 




9723 


5-i5i 


4.572 


9- 42^ 


ti 


34700 


67340 


73790 


35800 


58.8 




10. 112 


5.618 


4-494 


9425 


" 


35000 


63000 


78750 


37550 


59-3 




10.329 


5-745 


4.584 


9-425 


41 


36780 


66130 


82870 


40310* 


62.3 




10.624 


6.154 


4.470 


9-425 


44 


38120 


65810 


90600 


42970* 


64.6 




10.488 


6.078 


4.410 


9-425 


" 


34000 


58670 


80860 


37830 


57.6 




9- 2 33 


4-353 


4.880 


11.928 


44 


31050 


65860 


58750 


24030 


52.6 




, 9-596 


4.520 


5.076 


11.928 


•' 


32000 


67940 


65340 


25740 


54-2 




9.951 


4.983 


4.968 


11.928 


44 


34270 


68430 


68640 


28590 


58-o 




10.179 


5 082 


5-090 


11.928 


44 


33770 


67540 


67520 


28810 


57-2 




10 845 


5715 


5-130 


ii. 928 


■• 


34900 


66230 


7378o 


3*730 


59-i 




10.838 


5.708 


5-130 


11.928 


44 


358io 


67990 


75650 


32540 


60.6 




"-I75 


6.146 


5.029 


11.928 


60420 


38470 


69940 


85480 


36040 


63.6 




10.890 


5-996 


4.894 


11.928 


" 


3774o 


68650 


83980 


34460 


62.4 




9.624 


5-5or 


4.123 


9.940 


59000 


35000 


61230 


81700 


33890 


57-0 




9.776 


5.572 


4.204 


10.030 


44 


36470 


63990 


84810 


3555o 


61.7 




TO. 478 


6.192 


4.286 


9.940 


ti 


38760 


65590 


94750 


40850* 


65.7 




IO.OO4 


5-915 


4.089 


9.940 


44 


36740 


62130 


89880 


36970 


62.2 




IO.39O 


6329 


4.061 


9.940 


i< 


3793° 


62270 


97050 


39650* 


64.3 




IO.663 


6-495 


4.168 


9.940 


M 


40630 


65810 


90600 


42970* 


68.8 





* Steel rivets. 



62 2 



APPLIED MECHANICS. 



TABULATION OF SINGLE- 
STEEL PLATE— Continued. 



No. 

of 

Test. 


Sheet Letters. 


Pitch. 


No. 

of 

Rivets. 


Width 

of 
Joint. 


Nominal 
Thickness. 


Size of 
Rivets 

and 
Holes. 


Actual 
Thick- 
ness of 
Plate. 


Lap. 


Plate. 


Covers. 


Plate. 


Covers. 










in. 




in. 


in. 


in. 


in. 


in. 


in. 


ib°4 


Z 


M 


L 


2* 


6 


i3-5o 


3/4 


7/16 


i& * ii 


.722 


2 


*5°5 


Z 


M 


M 


it 


6 


" 


" 






.762 


2 


1506 


Y 


N 


O 


2| 


6 


14-25 


" 






•727 


2 


1507 


Y 


N 


O 


" 


6 




" 






•735 


2 


1508 


Y 


Q 


Q 


2* 


6 


15-00 


" 






•737 


2 


T 5°9 


Y 


Q 


Q 


" 


6 




" 






•753 


2 


1510 


Y 


L 


L 


2f 


5 


13-13 


" 






•748 


2 


15" 


Y 


L 


L 


" 


5 




" 






•755 


2 


T512 


Z 






2* 


5 


13-75 


" 






.750 


2 


1513 


Z 


N 


N 


" 


5 




M 






.764 


2 


1514 


Y 


O 





2| 


5 


I4-I3 


it 






.760 


2 


1515 


Y 


O 


O 


" 


S 




" 






.746 


2 


1516 




Pi 


P 


3 


5 


15.00 


" 






•749 


2 


1517 


Y 


^ 


Q 


»• 


5 




" 






.741 


2 


1518 


Z 


K 


K 


3i 


4 


12.50 


" 






•756 


2 


1519 


Z 


K 


K 


" 


4 




(i 






.741 


2 


1520 


Z 


K 


K 


3i 


4 


13.00 


" 






•763 


2 


1521 


Z 


K 


K 


" 


4 




" 






.718 


2 


1522 


Z 


M 


M 


3t 


4 


I3-50 


" 






•742 


2 


1523 


Z 


M 


M 


" 


4 










•754 


2 



TABULATION OF RIVETED JOINTS 



623 



RIVETED BUTT-JOINTS— Continued. 

STEEL PLATE -Continued. 



, Sectional Area 
of Plate. 


Bearing 
Surface 

of 
Rivets. 


Shear- 
ing 
Area of 
Rivets. 


Tensile 
Str'gth 

of Plate 

per 
Sq. In. 


Max. Stress on Joint per Sq. In. 


*o 

V 

pa 




Tension 
on Gross 
Stction 
of Plate. 


Tension 
on Net 
Section 
of Plate. 


Comp. on 
Bearing 
Surface 

of Rivets. 


be 


Gross. 


Net. 


sq. in. 


sq. in. ; 


sq. in. 


sq. in. 


lbs. 


lbs. 


lbs. 


lbs. 


lbs. 






c.761 


4.346 i 


5-415 


14.726 


59000 


29090 


65350 


52460 


19280 


49-3 




10.287 


4 572 


5-715 


14.726 


59000 


30010 


67520 


54010 


20960 


50.8 




10.367 


4.914 


5-453 


14.726 


60420 


33610 


70900 


^3890 


23660 


55-6 




10.474 


4.961 


5-5I3 


14.726 


60420 


33660 


71070 


63960 


23940 


55-7 




11 .070 


5-542 


5.528 


14.726 


60420 


34780 


69470 


69650 


26140 


57-5 




n. 310 


5.662 


5-6 4 8 


14.726 


60420 


34670 


69250 


69420 


26620 


57-4 




9.918 


5-243 


5-675 


12.272 


60420 


36120 


68380 


76680 


29210 


59-7 




9.928 


5-209 


4.719 


12.272 


60420 


36940 


70400 


77710 


29880 


61. 1 




10.328 


5.640 


4.688 


12.272 


59000 


3373o 


61770 


74320 


28390 


57.0 




10 505 


5-73° 


4-775 


12.272 


59000 


35260 


64640 


7757o 


30100 


59-7 




10.929 


6.179 


4-750 


12.272 


60420 


3793o 


67080 


87260 


36220 


62.7 




10. 735 


6.072 


4.663 


12.272 


60420 


38720 


68460 


89150 


33870 


64.0 




1 r . 205 


6.524 


4.681 


12.272 


55520 


36530 


62740 


87440 


36610 


65.8 




11. 108 


6-477 


4-631 


12.272 


60430 


38740 


66440 


92920 


35060 


64.1 




9-465 


5685 


3.780 


9.818 


59000 


37560 


62360 


93780 


36110 


636 




9-277 


5-572 


3-7°5 


9.818 


59000 


39000 


64930 


97650 


36850* 


66.1 




9-934 


6. 119 


3-815 


9.818 


59000 


37600 


61040 


97900 


38040* 


63-7 




9 348 


5-758 


3 590 


9.818 


59000 


36000 


58440 


9374o 


34280 


61.0 




10.032 


6.322 


3-71° 


9.818 


59000 


40040 


63540 


108270 


40910* 


67.7 




10.187 


6.417 


3-770 


9.818 


59000 


39720 


63050 


107320 


41210* 


67-3 





Steel rivets. 



624 



APPLIED MECHANICS. 



TABLE 
Table of Efficiencies of 



STEEL PLAT!. 







Pitch of Rivets. 




Plate. 


No. of 
Test. 




















1*" 


if" 


if" 


2" 


2*" 






per cent. 


per cent. 


per cent. 


per cent. 


per cent. 




1308 


67-5 


68.9 


72.8 








1313 


68.3 


69.9 


72.1 








13H 




65-9 


66.4 


68! 3 


68.7 


i" . . . .- 


1323 




64.8 


67.1 


68.6 


70.3 


1324 






62.6 


64.4 


67-5 




1337 






63-9 


65-9 


68.2 




1338 








64.8 


61.7 




1355 

1356 
1359 


74.6 
72.7 


73.3 
72.0 




58.7 


60.6 


f" . . . . • 


1360 
1367 




68.2 
68.3 


69.0 
70.1 


70.0 
7°-3 


iii 


1368 






64-5 


61.7 


64.8 




1379 






63.0 


65.2 


67.4 




1380 








58.7 


59-6 




1395 








57-3 


59-5 








200 


soo 








1396 




66.1 


79.6 


64.7 






1401 




68.1 


68.6 


350 
68.9 






1402 




.... 


64-5 

260 


66.3 
300 


400 

76.0 


*"....- 


1411 
1412 


.... 




66.9 


76.4 
250 
62.8 


66.3 
59-2 

800 




1425 




.... 


.... 


60.I 


68.1 

300 




1426 










59-3 


• 


1443 










5i.5 

300 




1444 


.... 


.... 


60.I 


66.0 


76.2 


*"....- 


I45 1 






62.7 


63.8 


66.9 


1452 








58.3 


60.0 




1463 








58.3 


61.8 




1464 


..,. 








56.5 




1481 


.... 








56.5 




1482 


.... 






54-2 


58.3 




1489 


.... 






54-2 


58.8 


*" . . . 


1490 






.... 




52.6 


1503 










54-2 




1504 














1523 













Notes.— Figures in heavy- face type denote that 
Super numbers state the temperature of 



TABULATION OF RIVETED JOINTS. 



625 



KiO. 1. 

Single-riveted Butt-Joints. 











STEEL PLATE. 
















Pitch of Rivets. 




Diam- 














eter of 
Rivet- 
















| 




*" 


2f" 


2±" 


2|" 


2f" 


2*" 


3" 


3i" 


3i" 


3*" 


holes. 


per ct. 


per ct. 


per ct. 


per ct. 


per ct. 


per ct. 


per ct. 


per ct. 


per ct. 


perct. 


in. 
I 


72.7 
















.... 




* 


74.0 




















* 


70.1 


70.7 


76. 8 


69^8 














* 


70.0 


72-3 


67.1 


68.3 










.... 


.... 


i 


61.5 


637 
68.4 


72.1 


65 -4 


77.1 


735 


m 








1 


63.4 


68.5 


68.4 


75.o 


74.1 








1 






















# 






















1 


















... . 




* 










— 












i 


70.7 


72.4 


75-5 
















i 


68.2 


70.4 


75-6 
















1 


65-4 


65-3 


69.8 


67-4 


68!8 


73-2 










1 


64.1 


68.3 

360 


69.5 


68.3 


69.8 


75-o 




.... 






1 

i 
i 


66.4 


75.5 
















.... 


s 


600 






















77.6 


66.7 


















$ 


200 


400 


600 




600 














60.3 


72.8 


77-3 


67.2 


81.5 












z 


6l.5 


64-5 


66.9 


400 
76.9 


67.6 












z 


360 


700 






600 


600 


200 










67.2 


62.7 


62.3 


62.O 


70.5 


80.3 


67.5 


67.2 






I* 


56.5 


58.0 


600 

76.7 


6l.2 


6 4 -5 


67.O 


70.0 


68.2 






It 


400 






















68.6 




















i 


600 

643 
63.6 




















* 

z 


65*8 


66.6 


65.0 














62.8 


65-5 


66.1 


64.5 














I 


57-8 


59.8 


62.5 


66.1 


66^5 


6 7 :i 


68.7 


67.9 






It 


58.0 


60.7 


62.1 


64.7 


66.5 


66.6 


68.6 


69.9 






It 


59.3 


64.6 


















z 


62.3 


57.6 


















I 


58.0 


59-i 


63.6 


67.0 


65"? 


64*3 










It 


57-2 


60.6 


62.4 


61.7 


62.2 


68.8 










1* 


49-3 


55 6 


57-5 


59-7 


57.0 


62.7 


65.8 
64.1 


63.6 


63 -7 
61.0 


67.'7 


li 


50.8 


S5 7 


57-4 


61. 1 


59 7 


64.0 


66.1 


67.3 


I* 



joint did not fracture along- line ol riveting, 
'oints tested at temperatures above atmospheric. 



626 



APPLIED MECHANICS. 



TABLE NO. 2. 

Table of Differences between the Efficiencies and Ratios of Net to 

Gross Areas. — Single-riveted Butt-Joints, Steel Plate. 









Width of Plate between Rivet Holes. 


Diameter 


Plate. 


No. of 










of Rivet 












Test. 










,; 1 „' 








Holes. 






1" 


if" 


if" 


if" 


if" if" 


if" 


if" 


2" 




in. 


in. 


perct. 


per ct. 


perct. 


perct. 


per ct. 


1 
perct. 


per ct. 


perct. 


per ct. 


in. 




1308 
1313 


6.0 
6.8 


4.6 
5-6 


6.1 
5-4 














f 
f 




13U 


8.8 


6-3 


7.2 


4.0 


7-5 










i 


i ■ 


1323 


7-7 


7.0 


6.1 


5-6 


8.8 










i 


1324 


8.0 


8.1 


8.6 


11. 7 


7.6 


ii.'8 


3-i 






i 




1337 


9-3 


9.6 


9.4 


8.9 


9.2 


2.1 


1.6 






f 




1338 


14.8 


8.8 


5-9 


5-8 


12. 1 


3-5 


13-4 


'8. '2 


£? 


1 




1355 


18.8 


7.6 


7.8 


10.5 


8-5 


6.6 


11. 3 


8.8 


1 


r 


1356 
1359 


*3-i 
11. 2 


9.1 

7.7 
















f 

* 




1360 


11. 1 


9.0 


7-5 


*6.' 2 












i 


t - 


1367 


11. 2 


10. 1 


7-8 


3.8 












i 


1368 


11. 2 


5-4 


5-9 


9-6 


9.4 


10.5 








f 




1379 


9-7 


8.9 


8.6 


7-i 


7.2 


10.6 








i 




1380 


6.8 


6.7 


9-9 


7-4 


9.8 


5-5 


5-2 


7-9 




1 




1395 


7-3 


6.6 


5-9 


10.4 


9-5 


6.4 


6.2 


9.8 




z 


r 


1396 
1401 


200 
9.0 

11. 1 


300 
19.6 

8.6 


2.1 

350 

6.4 
400 




l! 5 .°3 










1 
i 




1402 


11. 1 


10.0 


17.2 


5-3 










i 






250 


300 




500 














i ■ 


1411 


13-5 


20.I 


7-4 


16.4 


3.5 










1 




250 




200 


400 


500 




500 










1412 


12.8 


6.2 
300 


4-7 


I4.9 


I7.2 


5-2 

400 


17.8 






1 




1425 


IO. I 


15-2 


5-9 


6-5 


6.9 


I4.9 


4.0 






1 






300 


350 


700 






600 


500 


200 








1426 


12.2 


17.2 


I0.2 


6-9 


4-9 


II. 4 


19.4 


5-o 


3.2 


if 


i 


1443 


4.6 


6-5 


5-3 

300 


500 
21.7 

400 


4-i 


5-4 


6.1 


7-5 


4.2 


-if 




1444 


6.4 


9.7 


I7.4 


7-5 












i 


# - 


145 1 


9-4 


7-5 


8.1 


600 
3-2 












f 


1452 


8-3 


7-i 


8.0 


8.1 


6.6 


3."i 












1463 


8-3 


8.9 


7.2 


7.6 


6.0 


2.7 












1464 


9-5 


7.8 


7.2 


7-5 


9.0 


7-5 


'6.2 


6.2 


3.9 


if 


i 


1481 


9.4 


8.0 


8.0 7.1 


7.6 


7-4 


5-8 


6.1 


5.9 


if 




1482 


4-i 


5-3 


3.7 


6.7 
















1489 


4-i 


5 8 


6.6 


-0.4 
















1490 


5-5 


7-9 


6.4 


8.6 


-.02 


"6.6 


3-4 ' 






if 




1503 


7-i 


7.2 


7-9 


7-3 


4-7 


3.1 


7-9 






if 




1504 


4.8 


8.2 


7-4 


6.8 


2.4 


6.2 


H 


3.5 


2.1 


if 


* 


1521 


6.4 


8-3 


7-3 


8.6 


5-i 


7-4 


6.0 


-0.6 


if 




2f" 


per ct. 






















1.7 




• 




















4.3 


1 



Notes. — Figures in heavy-faced type denote that joint did not fracture along line of riveting 
Super numbers state the temperature of joints tested at temperatures above 
atmospheric. 



TABULATION OF RIVETED JOINTS. 



627 



TABLE NO. 3. 

Excess in Strength of Net Section in Joint over Strength of Tensile 
Test-Strip.'— Single-riveted Butt- Joints, Steel Plate. 









Width of Plate between Rivet Holes. 




Diameter 


Plate. 


No. of 








of Rivet 




« 






!Test. 




















Holes. 






1" 


ii" 


ii" 


if" 


H" . 


i£" 


ii" 


ii" 


2" 




in. 




per ct. 


perct. 


perct. 


per ct. 


per ct. 


per ct. 


per ct. 


per ct. 


perct. 


in. 


■ 


1308 
1313 


9.8 
11. 1 


7.2 
8.8 


9.1 
8.1 














* 

* 




1314 


i5-4 


10.5 


9.2 


6.2 


9-i 










* 


i ■ 


x 323 


i3-4 


11. 7 


9-7 


8.5 


11. 










I 


i3 2 4 


17.4 


14.4 


147 


14.7 


11. 9 


16. 1 


4-7 






i 




J 337 


19.7 


17.0 


15-9 


14.6 


14.4 


3-2 


2-5 






1 




1338 


29.6 


16.5 


10.6 


10. 


20.1 


5-6 


21.0 


12.7 


10.6 


1 




1355 


17.6 


T 4-5 


14.2 


18.0 


14. 1 


10.7 


17.7 


13-5 


10.1 


1 


f 


I3S6 


21.2 


14.1 
















s 




1359 


18. 1 


11.9 
















* 


1 


1360 


i9-3 


15-0 


12.0 


H 












i 


M 


i3 6 7 


19-5 


16.8 


12.5 


5.8 












i 


1368 


20.9 


9-7 


10. 1 


15.6 


14.9 


io!i 








i 




1379 


18.2 


15-9 


13-5 


11. 7 


11. 4 


16.5 








1 




1380 


15-5 


12.6 


17.8 


12.8 


16.3 


8.9 


'k'.2 


12.2 




1 


I 


1395 


i 4 .6 
200 


12.5 

300 


10.6 


17.9 


15-9 


10.3 


9-8 


*5-i 




z 




1396 


15-6 


32.8 


3-4 














* 




1401 


15.8 


14.4 


360 

10.3 














1 




1402 


20.9 

350 


i 7 .8 

300 


400 
28.9 


8.7 

600 


l 3 9 6 .°5 








.... 


t 


* 


141 1 


25.2 


35-8 


12.8 


27.O 


5.6 










t 




250 




200 


400 


600 




500 










1412 


25-5 


11. 8 

300 


5-5 


25.8 


28.7 


8.6 

400 


28.1 






1 




1425 


20.3 


28.8 


10.8 


"•3 


"•3 


24.O 


6.4 






I 






soo 


360 


700 






600 


500 


200 








1426 


26.0 


34-4 


12.3 


12. 1 

600 


8.7 


I9.4 


32.O 


8.1 


5.3 


I* 


1 


1443 


9.8 


13.2 


10. 1 

300 


39-7 

400 


7-3 


9.1 


IO. I 


12.2 


6.5 


Ii 




1444 


i5-i 


18.8 


29.6 


12.2 
600 












* 


t ■ 


1451 


17.7 


13-5 


13.7 


5-2 












i 


1452 


16.6 


17.2 


14.6 


14.O 


li'.i 


5.2 








I 




1463 


16.6 


16.8 


13.2 


I3.2 


10.1 


4.3 








I 




1464 


20.2 


15-8 


13.6 


13-7 


157 


12.8 


I0.2 


io.i 


6.'i 


Ii 


. 


1481 


20.0 


18. 4 


15-3 


I3.0 


13-5 


12.7 


9-7 


9.8 


9.2 


Ii 




1482 


8-3 


10. 1 


6.8 


"•5 












I 




1489 


8.2 


14.1 


12. 1 


-0.6 












I 




1490 


n. 6 


16.0 


12.3 


15-8 


3.8 


11. 2 


5-5 






Ii 




1503 


15.2 


14-5 


152 


13.6 


8.5 


5.3 


"•5 






Ii 




IS04 


10.8 


17.4 


15.0 


13.2 


4-7 


11. 


130 


5.' 7 


3-5 


Ii 


i ■ 


1521 


14.4 


17.6 


14.6 


16.5 


9-6 


13-3 


10.0 


IO.I 


-0.9 


Ii 
Ii 


2i" 


per ct. 






















7.7 


Ii 


. 




















6.9 


Ii 


Average of 




















all joints. 


16.2 


14.4 j 12.3 


12.9 


11. 9 


11. 


9-7 


11. 8 


7.0 





Notis.— Figures in heavy-faced type denote that joint did not fracture along line of riveting. 
Super numbers state the temperature of joints tested at temperatures above 
atmospheric. 



628 



APPLIED MECHANICS. 



In the Report of Tests made at Watertown Arsenal during the fiscal year 
ended June 30, 1891, is the following account of another series of tests on riveted 
joints: 

"Comprised in the present report are 113 tests made with steel plates of 
1/4", 5/16", 3/8", and 7/16" thickness with iron rivets machine driven in drilled 
or punched holes. 

"The plates used were from material used in earlier tests, the results of 
which have been published in previous reports. 

" In the use of metal once before tested, such plates were selected as had 
not been overstrained previously, or those in which the elastic limit had been 
but very slightly exceeded. 



SINGLE-RIVETED 



STEEL PLATE. 



No. of 
Test. 


Sheet Letters. 


Pitch. 


No. of 
Rivets. 


Width 

of 
Joint. 


Nominal 
Thickness. 


Size 

and 
Kind 

of 
Holes. 


Actual 
Thick- 
ness 
of 
Plate. 


Lap. 


Plate. 


Covers. 


Plate. 


Covers. 










in. 




in. 


in. 


in. 


in. 


in. 


in. 


4913 


H 


C 


D 


2* 


5 


13-72 


1/4 


3/16 


7/8 d 


• 247 


2 


4914 


L 


D 


D 


" 


5 


13.69 


" 




M i< 


.248 


" 


49i5 


M 


E 


E 


2| 


5 


14-32 


it 




ti (I 


.247 


" 


4916 


M 


D 


D 


" 


5 


M -33 


M 




It 11 


.247 


" 


4917 


M 


E 


E 


3 


5 


15.00 


11 




" " 


.246 


it 


4918 


M 


E 


E 


" 


5 


14.98 


" 




It u 


.247 


44 


498s 


Q 


D 


D 


3i 


4 


14.00 


ft 




I " 


• 3°9 


1 


4987 


s 


C 




u 


4 


14.01 


(t 




" " 


.310 


x x 


499i 


Q 


■ 




u 


4 


14.05 


«< 




11 (« 


.308 


i* 


5*25 


R 


A 


A 


1 


10 


10.02 


5/l6 




1/2 " 


.306 


ii 


5126 


R 


A 




i* 


8 


10.02 


" 




(( II 


-3°4 


44 


5127 


R 


B' 




v ** 


7 


10.51 


" 




II 1. 


.310 


" 


5143 


L 


P 




2 


7 


14.03 


7 /i6 


5/16 


7/8 P . 


.440 


ii 


5i44 


L 


O 


O 


" 


7 


14.01 


re 




" d. 


.440 


" 


5145 


O 


Q 


Q 


2} 


6 


i3-5o 


it 




" " 


•434 


" 


5146 


M 







" 


6 


13-Si 


" 




" P- 


.421 


" 


5147 


O 


P 


p 


2* 


6 


15.02 


" 




M d. 


.413 


44 


5148 


N 


P 




" 


6 


15.02 


" 




" P- 


.411 


" 


5i55 


K 


s 


.... 


2* 


5 


13-75 






M d. 


.425 





TABULATION OF RIVETED JOINTS. 



629 



"The present tests are supplementary to those of earlier reports, and occupy 
a place intermediate between the elementary forms of joints and the more elab- 
orate types of joints which have been investigated. 

" Wide variation has been given the pitches, and rivets of extreme diameters 
have been used for the purpose of including joints in which these features have 
been carried to their extreme limits. 

" The efficiencies of the joints are stated in per cent of strength of the solid 
plate." 



BUTT-JOINTS. 



STEEL PLATE. 



Section 
of P 

Gross. 


al Area 
late. 

Net. 


Bearing 
Surface 

of 
Rivets. 


Shear- 
ing 
Area 
of 

Rivets. 


Tensile 
Str'gth 

Plate 

per 

Sq. In, 


Maximum Stress on Joint per Sq. In. 


Effi- 
ciency 

of 
Joint. 


Tension 

on 

Gross 

Section 

of Plate. 


Tension 
on Net 
Section 

of 
Plate. 


Compres- 
sion on 
Bearing 
Surface 

of Rivets. 


Shearing 

of 
Rivets. 


sq. in. 


sq. in. 


sq. in. 


sq. in. 


lbs. 


lbs. 


lbs. 


lbs. 


lbs. 




3-39 


2.31 


1.08 


6.01 


59180 


44180 


65760 


140650 


25270 


75-7 


3-4° 


2.31 


1.09 


6.01 


61470 


43500 


64030 


135690 


24610 


70.8 


3-54 


2.46 


1.08 


6.01 


58170 


46300 


66630 


151780 


27270 


79.6 


3-54 


2.46 


1.08 


6.01 


58170 


43500 


62590 


142570 


25620 


74.8 


3-69 


2.61 


1.08 


6.01 


58170 


46290 


65440 


158150 


28420 


79.6 


3 -70 


2.62 


1.08 


6.01 


58170 


44400 


62700 


152110 


27330 


76-3 


4-33 


3-°9 


1.24 


6.28 


56760 


24040 


33690 


83950 


16580 


42.3 


4-34 


3.10 


1.23 


6.28 


57000 


26770 


37480 


93710 


18500 


46.9 


4-33 


3.10 


1.23 


6.28 


56760 


33940 


47410 


119500 


23400 


59-8 


3.07 


i.54 


i-53 


3-92 


61130 


35930 


71620 


72090 


28.140 


58.8 


3 -°5 


1.83 


1.22 


3.14 


61130 


41280 


68800 


103200 


40100 


67.5 


3.26 


2.17 


1.09 


2.74 


61130 


39250 


58960 


1 1 7380 


46690 


64.2 


6.17 


3.38 


2.79 


8.41 


59390 


32540 


59410 


71970 


23880 


54-8 


6.16 


3.48 


2.69 


8.41 


5939° 


23360 


41350 


5349o 


17110 


39-T i 


5-86 


3-58 


2.28 


7.21 


52910 


42250 


60160 


108600 


3434o 


79.8 


5.69 


3-4° 


2.29 


7.21 


61650 


3974o 


66500 


98730 


31360 


64-5 


6.20 


4-03 


2.17 


7.21 


52910 


44*50 


67920 


126130 


37960 


83.4 


6.17 


3-94 


2.23 


7.21 


61650 


36660 


57410 


101430 


31370 


60.0 


5-84 


3.98 


1.86 


6.01 


59000 


40270 


59100 


126450 


39130 


68.3 



630 



APPLIED MECHANICS, 



TABULATION OF* SINGLE- 



STEEL PLATE. 



No. of 
Test. 


Sheet Letters. 


Pitch. 


No. of 
Rivets. 


Width 

of 
Joint. 


Nominal 
Thickness. 


Size and 
Kind 

of 
Holes. 


Actual 
Thick- 
ness of 
Plate. 


Lap. 
in. 


Plate. 


Plate. 


Plate. 


Plate. 








in. 




in. 


in. 


in. 


in. 


in. 


4933 


I 


J 


2* 


5 


10.62 


1/4 


1/4 


fd. 


.252 


2 


4934 


J 


J 


" 


5 


10.65 


" 




" 


-253 


2 


4939 


L 


K 


2| 


4 


11.50 


" 


" 


1 d. 


.250 


2 


4940* 


J 


J 


" 


4 


11.50 


" 




" 


.256 


2 


494i 


K 


J 


tt 


4 


11 51 


" 




ifd. 


.252 


2 


4942 


K 


J 


11 


4 


11.52 


11 




" 


.250 


2 


4943 


K 


K 


" 


4 


11.50 


" 




ifd. 


.252 


2 


4944 


E 


J 


it 


4 


11.50 


it 




it 


•253 


2 


4945 


K 


K 


3* 


4 


12.52 


" 




11 


.248 


2 


4946 


L 


G 


" 


4 


12.55 


11 




11 


•253 


2 


4947+ 


N 


H 


" 


4 


13-52 


" 




it 


.247 


2 


4948$ 


N 


H 


" 


4 


13-52 


11 




" 


.247 


2 


4949 


M 


L 


3# 


4 


14.51 


" 




" 


.248 


2 


4950* 


M 


L 


ii 


4 


14.51 


it 




it 


.247 


2 


4961 


B 


E 


if 


6 


10.52 


3/8 


3/8 


fd. 


.388 


2 


4979 


E 


E 


2f 


5 


11.84 


" 




1 d. 


.384 


2 


5131 


K 


K 


if 


8 


12.00 


7/16 


7/16 


Id. 


• 427 


i-75 


5132 


N 


O 


" 


8 


12.00 


" 




fp. 


•,4i5 


i-75 


5X33* 


K 


K 


If 


8 


13.00 


11 




id. 


•427 


i-75 


5i34 


N 


N 


." 


8 


13.00 


" 




fP- 


•413 


i-75 


5*35 


M 


M 


If 


8 


14.03 


M 




*d. 


.422 


1-75 


5136 






" 


8 


13-99 


ii 




IP- 


.420 


i-75 


5137 


L 


M 


2 


7 


14.02 


" 




fd. 


•4«4 


i-75 


5138 


P 


M 


" 


7 


14.05 


it 




fp. 


.420 


i-75 


5139 


O 


K 


II 


6 


12.06 


" 




ifd. 


.428 


2 


5HO 


M 


M 


2f 


6 


14.28 


" 




•4 


.421 


2 


5i4i* 


L 


L 


2f 


5 


13-73 


it 




« 


.438 


2 


5142* 


Q 


Q 


3i 


5 


15-67 


11 




If 


.422 


2 



* Pulled off rivet-heads. 
t Pulled off 3 rivet-heads. 
t Pulled off 2 rivet-heads. 



TABULATION OF RIVETED JOINTS. 



631 



RIVETED LAP-JOINTS. 



STEEL PLATE. 



Sectional Area 
of Plate. 


Bear- 
ing 
Surface 

of 
Rivets. 


Shear- 
ing 
Area 
of 
Rivets. 


Tensile 

Strength 
of Plate 

per 
Sq. In. 


Maximum Stress on Joint per 


Sq. In. 


Effi- 
ciency 

of 
Joint. 


Tension 
on Gross 

Section 
of Plate. 


Tension 
on Net 
Section 
of Plate. 


Comp. on 
Bearing 
Surface 

of Rivets. 


Shear- 
ing of 
Rivets. 


Gross. 


Net. 


sq. in. 


sq. in. 


sq. in. 


sq. in. 


lbs. 


lbs. 


lbs. 


lbs. 


lbs. 




2.68 


i-57 


1. 10 


3-oo 


61000 


3975o 


67850 


96840 


35510 


65.1 


2.70 


i-59 


1. 11 


3-oo 


61000 


39660 


67360 


96490 


35700 


65.0 


2.87 


i.8 7 


1. 00 


3.14 


58150 


40560 


62250 


1 16400 


37070 


69.7 


2.94 


1.92 


1.02 


3-i4 


61000 


37010 


56670 


106670 


34650 


60.6 


2.90 


1.77 


1-13 


3.98 


61000 


43280 


70900 


1 1 1060 


31530 


70.9 


2.88 


i-75 


*«*3 


3.98 


58150 


42770 


73880 


119010 


30950 


73-5 


2.90 


1.64 


1.26 


4.91 


58150 


41130 


72730 


94660 


24290 


70.7 


2.91 


1.64 


1.27 


4.91 


58150 


40200 


71330 


92110 


23820 


69.1 


3.10 


1.86 


1.24 


4.91 


58150 


40030 


66720 


100080 


25270 


69.1 


3-i7 


1. 91 


1 26 


4.91 


61470 


41770 


69320 


105080 


26970 


68.0 


3-34 


2.10 


1.24 


4.91 


5574o 


42240 


67180 


112970 


28730 


75-7 


3-34 


2.10 


1.24 


4.91 


59180 


42600 


67760 


114760 


28980 


71.9 


3.60 


2.36 


1.24 


4.91 


61470 


41390 


63140 


120180 


30350 


67-3 


3-58 


2-35 


1.23 


4.91 


58170 


42150 


64210 


122680 


30730 


72.4 


4.08 


2.33 


*-7S 


2.65 


58340 


25950 


4544o 


60500 


39950 


44.4 


4-55 


2.63 


1.92 


3-93 


58340 


33050 


57190 


78330 


38270 


56.6 


512 


2.13 


2.99 


4.81 


59000 


31740 


76290 


5435o 


33780 


53-8 


4 99 


1.98 


3.01 


4.81 


52910 


27820 


70100 


461 10 


28860 


52.6 


5-55 


2.56 


2.99 


4.81 


59000 


31100 


67420 


57730 


35880 


52.7 


5-37 


2-37 


2.99 


4.81 


61 140 


30370 


68820 


5455o 


33910 


49-7 


5-9° 


2-95 


2.95 


4.81 


61650 


29240 


58490 


58490 


35870 


47-4 


5-89 


2-95 


2.94 


4.81 




31870 


63630 


63840 


39020 




5-94 


3-35 


2.60 


4.21 


5939o 


27580 


48900 


63000 


38910 


46.4 


5-9° 


3-24 


2.66 


4.21 


52910 


28530 


51940 


63270 


39980 


53-9 


5.16 


i-95 


3.21 


7.36 


58090 


28190 


74610 


45320 


19770 


48.5 


6.01 


2.85 


3.16 


7.36 


61650 


34850 


7349o 


66280 


28460 


56.5 


6.01 


3.28 


2.74 


6.14 


5939o 


33560 


61490 


73610 


32850 


56.5 


6.61 


3-97 


2.64 


6.14 


56960 


30420 


50650 


76170 


32750 


53-4 



632 



APPLIED MECHANICS. 



TABULATION OF DOUBLE- 

CHAIN-RIVETING-STEEL PLATE. 



No. 

of 

Test. 


Sheet Letters. 


Pitch. 


o" 
" i: 

<u . 


Total 
Num- 
ber of 
Riv- 
ets. 


Width 

of 
Joint. 


Nominal 
Thickness. 


Size 

and 

Kind of 

Holes. 


Actual 
Thick- 
ness ol 
Plate. 


Lap. 












Plate. 


Covers. 




5 <u 
Q 






Plate. 


Covers . 
















in. 


in. 




in. 


in. 


in. 


in. 


in. 


in. 


4911 


K 


C 


C 


2* 


2* 


10 


13.10 


1/4 


3/16 


S/8d. 


.253 


ill 


4912 


L 


C 


c 


" 


" 


10 


13.10 


it 


(i 


" 


• 253 


" 


4919 


L 


E 


D 


2* 


2* 


10 


14.32 


" 


« 


7/8 d. 


.247 


ill 


4920 


L 


E 


D 


" 


*' 


10 


14.32 


" 


(« 


" 


•249 


" 


4921 


K 


C 


B 


3* 


" 


8 


12.52 


" 


11 


" 


.252 


M 


4922 


K 


C 


C 


" 


" 


8 


12.49 


" 


" 


tt 


.252 


tt 


4923 


J 


A 


A 


3* 


" 


6 


n. S7 


" 


«i 


" 


• 257 


tt 


4924 


J 


A 




" 


" 


6 


"•53 


" 


i« 


tt 


.255 


" 


4925 


L 


B 




4l 


" 


6 


13.09 


" 


i« 


" 


.251 


" 


4926 


K 


C 


B 


" 


" 


6 


13.10 


" 


" 


it 


.230 


" 


5128 


R 


C 


G 


if 


2 


M 


12.27 


5/16 


3/x6 


1/2 d. 


.304 


if 


5129 


Q 


C 




2 


" 


14 


14.00 


" 


" 


" 


.305 


" 


5130 


s 


D 




2i 


" 


12 


13-58 


" 


" 


it 


• 307 


" 


4993 


Q 


E 




3* 


ii 


8 


14.05 


" 


" 


id. 


•309 


if 


4995 


Q 


E 




" 


if 


8 


14.06 


" 


" 


" 


•305 


1* 


4997 


Q 


E 




" 


2 


8 


14.08 


" 


" 


" 


.308 


if 


. 4951 


B 


I 


I 


2* 


2j 


8 


8.52 


3/8 


1/4 


3/4 d. 


• 392 


1* 


4952 


E 


R 




" 


" 


8 


8.51 


" 


5/6 


" 


• 383 


1. 


4953 


E 


R 




af 


" 


8 


10.51 


" 


" 


tt 


.388 


" 


4954 


E 


N 


H 




" 


8 


10.03 


" 


1/4 


" 


384 


. " 


4955 


E 


S 




3i 


" 


8 


12.50 


" 


5/i6 


" 


.383 


tt 


4957 


H 


O 


O 


3f 


" 


8 


14-51 


" 


41 


tt 


•369 


" 


4958 


H 


M 


M 


" 


" 


8 


14-52 


" 


1/4 


" 


.369 


" 


4959 


B 


S 


S 


4* 


" 


6 


12.42 


" 


5/i6 


it 


.388 


if 


4960 


E 


L 


N 




" 


6 


12.42 


" 


1/4 


11 


.384 


" 


4967 


C 


M 


M 


2* 


2* 


10 


14 38 


" 


" 


id. 


• 375 


2 


4969 


E 


M 




3l 


" 


8 


13-50 


" 


it 


" 


.382 


it 


4970 


F 






" 


" 


8 


13-58 


" 


5/16 


it 


.380 


tt 


4971 


J 


P 


P 


3l 


t( 


8 


15.46 


" 


" 


it 


•379 


it 


4973 


E 


S 


S 


4t 


" 


6 


13-50 


" 


" 


tt 


.385 


" 


4975 


I 


O 


P 


4i 


tl 


6 


14.65 


" 


" 


" 


• 373 


" 


4977 


J 


P 


P 


5f 


" 


6 


.16.08 


" 


" 


tt 


•379 


tt 


4956 


K 


N 


N 


3i 


2* 


8 


12.48 


7/16 


1/4 


3/4 d. 


.427 


i5 


4968 


N 


O 




H 


2* 


10 


14.41 




5/i6 


id. 


.409 


2 



TABULATION OF RIVETED JOINTS. 



633 



RIVETED BUTT-JOINTS. 

CHAIN-RIVETING— STEEL PLATE. 



Sectional Area 
of Plate. 


Bear- 
ing 
Surface 

of 
Rivets. 


Shear- 
ing 
Area 
of 
Rivets. 


Tensile 

Strength 

of 

Plate 

per 

Square 

Inch. 


Maximum Stress 


on Joint per Sq. In. 


Effi- 
ciency 

of 
Joint. 


Gross. 


Net. 


Tension 

on 

Gross 

Section 

of 
Plate. 


Tension 
on 
Net 

Section 

of 
Plate. 


Compres- 
sion on 
Bearing 
Surface 

of 
Rivets. 


Shearing 

on 
Rivets. 


sq. in. 


sq. in. 


sq. in. 


sq. in. 


lbs. 


lbs. 


lbs. 


lbs. 


lbs. 




3-3* 


2.52 


1.58 


6.14 


58150 


49090 


64480 


102850 


26470 


84.4 


3-31 


2.52 


1.58 


6.14 


61470 


51960 


68250 


108860 


28010 


84.5 


3-54 


2.46 


2.16 


12.02 


61470 


46810 


67370 


76720 


13790 


76.1 


3-57 


2.48 


2.18 


12.02 


61470 


45700 


65790 


74840 


i3S7o 


74-3 


3.16 


2.27 


1.76 


9.62 


58150 


46330 


64490 


83180 


15220 


79.6 


3*5 


2.27 


1.76 


9.62 


58150 


4 6 73o 


64850 


83640 


15300 


80.3 


2.97 


2.30 


x -35 


7.21 


61000 


49520 


63940 


108930 


20400 


81.2 


2.94 


2.27 


J -34 


7.21 


61000 


49460 


64050 


108510 


20170 


81. 1 


3-29 


2.63 


1.32 


7.21 


61470 


51440 


64350 


1282x0 


23470 


83-7 


301 


2.41 


1. 21 


7.21 


58150 


555oo 


69320 


138070 


23170 


95-4 


3-73 


2.66 


2.13 


5-49 


6 1 130 


46690 


66650 


83240 


32300 


76.4 


4.27 


3.20 


2.14 


5-49 


56760 


49040 


65430 


97850 


38140 


86.4 


4-i5 


3-23 


1.84 


4.70 


57000 


46480 


59720 


104840 


41040 


81.5 


4-34 


3-" 


2.47 


12.57 


56760 


4474o 


62430 


78600 


iS45o 


78.8 


4.29 


3-°7 


2.44 


12.57 


56760 


4549o 


63570 


79980 


15530 


80.1 


4-34 


3.10 


2.46 


12.57 


56760 


4553o 


63740 


80330 


15720 


80.2 


3-34 


2.16 


2-35 


7.07 


5973° 


43290 


66940 


61520 


20450 


72.4 


3.26 


2. 11 


2.30 


7.07 


58340 


42380 


65470 


60070 


19540 


72.6 


4,08 


2.91 


2-33 


7.07 


58340 


46590 


65330 


81590 


26890 


79.8 


3-85 


2.70 


2.30 


7.07 


58340 


49 x 3o 


70060 


82240 


26750 


84.2 


4-79 


3.64 


2.30 


7.07 


58340 


48610 


63970 


101250 


32940 


83.3 


5-35 


4-25 


2.21 


7.07 


56670 


48500 


61060 


1 1 7420 


36700 


8 5 -5 


S-36 


4-25 


2.21 


7.07 


56670 


47700 


60160 


1 15700 


36170 


84.1 


4.82 


3-95 


!-75 


5-30 


59730 


43070 


52560 


1 18630 


39170 


72.1 


4-77 


3-9i 


1-73 


5-30 


58340 


42520 


51870 


1 17230 


38260 


72.8 


5-39 


3-52 


3-75 


15-71 


57870 


42890 


65680 


61650 


14720 


74- 1 


5. 16 


3-63 


3.06 


12.57 


58340 


44263 


62920 


74640 


18170 


75-9 


S.16 


3- 6 4 


3-04 


12.57 


54290 


43240 


61290 


7339o 


i775o 


79.6 


5.86 


4-34 


3-03 


12.57 


57130 


44910 


60650 


86860 


20940 


78.6 


5.2° 


4.04 


2.31 


9.42 


58340 


45980 


59180 


103510 


25380 


78.8 


5-46 


4-35 


2.24 


9.42 


59030 


46720 


58640 


1 13880 


27080 


79.1 


6.09 


4.96 


2.27 


9.42 


S7130 


44650 


54830 


1 19800 


28870 


78.1 


5-33 


4 -05 


2.56 


7.07 


59000 


48120 


63300 


1 00190 


36280 


8 3 -3 


5-89 


3-85 


4.09 


I5-7I 


6 1 140 


43360 


66340 


62440 


16260 


70.9 



634 



APPLIED MECHANICS. 



TABULATION OF RIVETED 
DOUBLE-RIVETED LAP-JOINTS. 





S 8 " 




en qj 






















<*. V 




£.5 










Nominal 


<« 


10 




CO 

V 

H 


£ 




OJ 

*j Qficfl 
5 C tJ 


C 

go 




c 

g ° 


'0 


Thickness. 



•a 

c 

2 


u 

c 
J* 








o 


j rt 




& rt > 


«« 


££•0 


*T1 









•O oJ 


H-S 




o 






.2 rt 



dfe 


"S8 




-a 


to 
v 


u 
U 
> 


Si 


"rtS 
0*0 


a 




C/) 1 ^ 


s 


Q 


£ 


fc 


£ 


Pu, 


O 


w 


<J 


■J 






in. 


in. 








in. 


in. 


in. 


in. 


in. 


in. 


4935 


i-1 


H 


2| 


5 


5 




10.68 


i/4 




7/8 d. 


•257 


i« 


4936 






5 


5 




xo-53 


44 






.251 




4937 


L-M 


=4 


'" 


5 


5 




14.38 


41 




44 " 


.248 


44 


4938 


I-L 






5 


5 




14.40 


44 




it 11 


.249 


44 


4999 


M 


3i 


2i 


4 


4 




14.02 


5/i6 




1 ' 4 


.305 


2 


5000 






4 


4 




14.00 






1 p. 


.306 


44 


4963 


E-E 


1* 


" 


6 


6 




10.50 


3/8 




3/4 d. 


.387 


if 


496s 


E-E 


2 


" 


6 


6 




12.00 








.384 




498i 


E-D 


2* 


2j 


5 


5 




11.83 


41 




1 ' 4 


.38s 


2 


4983 


D-H 


2* 


44 


5 


5 




14.36 


44 




1 44 


•37° 


44 


5M9 


M-L 


2 


2| 


7 


7 




14.00 


7/16 




7/8" 


•425 


it 7 , 


5150 


M-M 


1 


" 


7 


7 




14.00 


44 




P- 


•423 




5151 


K-K 


2* 


" 


6 


6 




1 3-S3 


44 




" d. 


.428 


44 


5152 


L-O 


" 


" 


6 


6 




13-5° 


44 




" P- 


.440 


i* 


5i53 


O-O 


2* 


44 


6 


6 




15.01 


44 




41 d. 


.409 


x/s 


5154 


O-O 


" 


44 


6 


6 




15.02 


44 




" P- 


.412 




5156 


K-K 


2j 




5 


5 




13-77 






44 d. 


.422 





DOUBLE-RIVETED BUTT-JOINTS. 



4927 


MDE 


2* 


2* 


5 


4 




14.36 


1/4 


3/i6 


7/8 d. 


.250 


iM 


4928 


LDE 


44 


44 


5 


4 




J 4-35 








.247 




4929 


KBC 


a* 


44 


4 


3 




12.51 




44 




•255 




4930 


KC 


44 


4 


3 




12.50 


44 


44 


44 " 


.251 


44 


4931 


HDD 


4* 


44 


3 


2 




13.12 


44 


44 




.248 


44 


4932 


HCC 






3 


2 




13.12 








.246 





DOUBLE-RIVETED LAP-JOINTS. 



TREBLE-RIVETED LAP-JOINTS. 



5"9 


£P 


?* 


i* 


4 


3 




14.00 


5/^6 




1 d. 


•303 


*i 


5120 


44 


4 


3 




14.03 


" 




1 p. 


•305 


t4 


5121 


n 




1* 


4 


3 




14.03 






1 d. 


.302 


it 


5122 


" 


44 


4 


3 




14.03 


" 




1 p. 


•304 




5123 


0-0 




2 


4 


3 




14.02 


44 




1 d. 


.302 


x* 


5124 


p-p 






4 


3 




14.02 






1 p. 


.307 





5157 


KK 


2i 


?# 


. 5 


5 


5 


13-14 


7/16 




7/8 d. 


.432 


xtt 


5*58 


OP 


3 


44 


5 


5 


5 


15-05 


44 




i< it 


.412 


44 


5159 


PP 


3± 


44 


4 


4 


4 


12.78 


44 




it tt 


•432 


" 


5160 


LL 


3* 




4 


4 


4 


13-5° 








•43» 





TABULATION OF RIVETED JOINTS. 



635 



JOINTS.— STEEL PLATE. 



CHAIN-RIVETING. 



Sectional Area 
of Plate. 


V 
3 ui 


u 
<j 


rength 
per 


Maximum Stress on Joint per Sq. In. 


c 
'0 

*o 




















bCcn 
.5* 


£75 if . 


Tension 
on Gross 


Tension 
on Net 


Compres- 
sion on 


Shearing 
on 




c 


Gross. 


Net. 


Sis 


Section of 


Section 


Bearing 








<L> O 


Jo* 


£ OC/3 


Plate 


of Plate . 


Surface 


Rivets. 


1 






H 


C/2 


h 






of Rivets. 




W 


sq. in. 


sq. in. 


sq. in. 


sq. in. 


lbs. 


lbs. 


lbs. 


lbs. 


lbs. 




2.74 


1.62 


2.25 


6.01 


61000 


42770 


72350 


52090 


19500 


70.1 


2.64 


i-54 


2.20 


6.01 


62300 


42350 


72600 


58180 


18600 


67.9 


3-57 


2.49 


2.17 


6.01 


61470 


47870 


68630 


78760 


28440 


77.0 


3-58 


2.49 


2.18 


6.01 


61470 


48530 


69780 


79700 


28910 


78.9 


4.28 


3.06 


2.44 


6.28 


56760 


46070 


64440 


80820 




80.1 


4.28 


3-°3 


2-53 


6.28 


593°o 


43900 


62100 


7437o 


29960 


74.1 


4.06 


2.32 


3.48 


5-3o 


58340 


40570 


70900 


4733o 


-?io8o 


69-5 


4.61 


2.88 


346 


5-30 


58340 


42150 


67470 


56160 


36660 


72.2 


4-55 


2.63 


3-85 


7-8 5 


5373o 


38790 


67100 


45840 


22480 


72.2 


5-3i 


3-4 6 


3-7° 


7.85 


56670 


4^5° 


66070 


61780 


29120 


76.0 


5-95 


3-35 


5.21 


8.41 


61650 


40620 


72150 


46^90 


28740 


65.8 


5-93 


3-24 


5-36 


8. 4 r 


61650 


379 T o 


69380 


41940 


26730 


61.4 


5-79 


3-54 


4.49 


7.21 


59000 


43 x 5o 


70570 


55 f 4o 


34650 


73 - 1 


5-94 


3-55 


4.78 


7.21 


5939° 


38870 


65040 


48310 


?2020 


65 -4 


6.14 


3-99 


4.29 


7.21 


52910 


43530 


6f 990 


62310 


37070 


82.3 


6. tg 


3-95 


4.48 


7.21 


52910 


40380 


63290 


55800 


34670 


7 6 -3 


5.81 


3-96 


3-69 


6.01 


59000 


38850 


56990 


61170 


37550 


65.8 



ZIGZAG-RIVETING. 



3 59 


2.50 


1.97 


10.82 


58170 


48150 


69140 


87740 


15980 


80.3 


3-54 


2.46 


1.95 


10.82 


61470 


47420 


68240 


86090 


15520 


77.1 


3-19 


2.30 


1.56 


8.41 


58150 


46610 


64650 


95320 


17680 


80.2 


3- J 4 


2.26 


1-56 


8.41 


58150 


47520 


66020 


95640 


17740 


81.7 


3-25 


2.60 


1.05 


6.01 


59180 


47640 


59550 


147450 


25760 


80.5 


3- 2 3 


2.58 


1.08 


6.01 


59180 


46720 


5S490 


139720 


25110 


78.9 



ZIGZAG-RIVETING. 



CHAIN-RIVETING. 



4.24 


303 


2.12 


5-50 


56760 


42750 


59830 


85510 


32960 


75-3 


4.28 


3.02 


2 .20 


5 5° 


50300 


40630 


57580 


79050 
86180 


31620 


68. 5 


4-23 


3.02 


2-.IT 


5 50 


54350 


42990 


60220 


33060 


79.1 


4.27 


3 01 


2.19 


5-50 


5435o 


44340 


62000 


86450 


34420 


81 6 


4-23 


302 


2. II 


5-50 


54350 


44870 


62850 


89950 


345 ro 


82.5 


4-33 


303 


2.22 


5-5Q 


59300 


4349o 


62150 


84820 


31400 


73-3 



5-93 


4.04 


5-66 


9.02 


59000 


4=1720 


67100 


47900 


30060 


77 5 


6.20 


4.40 


5-4i 


9.02 


529IO 


48710 


68630 


55820 


33480 


92.1 


5-52 


4.01 


4-54 


7.21 


5809O 


48040 


66130 


58410 


36780 


82.7 


5.91 


4.38 


4 .6o 


7.21 


5939° 


46430 


62650 


59650 


38060 


78.2 



636 APPLIED MECHANICS. 

In the design of a riveted tension-joint the problem usually 
presents itself in the following form : 

Given, in all particulars, the two plates to be united, to 
design the joint ; i.e., to determine, i°, the diameter of rivet to 
be used ; 2°, the spacing of the rivets, centre to centre ; and, 3 , 
the lap. 

In regard to the determination of the lap, the common 
practice has been already explained and very little has been 
done experimentally. 

In order to determine the diameter and the spacing of the 
rivets by the usual methods of calculation, it becomes neces- 
sary to know the three following kinds of resistance of the 
metals, viz.: 

i°. The tensile strength per square inch of the plate along 
the line or lines of rivet-holes ; 

2°. The shearing-strength of the rivet metal ; 

3 . The resistance to compression on the bearing-surface of 
either plate or rivet. 

Hence we need to ascertain what the tests cited show in 
regard to these three quantities. 

Tension. — The tensile strength of the plate used should, 
of course, be determined by means of tests made on specimens 
cut from it. Further than this, questions arise as to the 
excess tenacity due to the grooved specimen form, and as to 
any injury due to punching when the holes are punched. 

The excess tenacity is, of course, greater with small than 
with large spaces between the rivet-holes ; hence, inasmuch as 
the tendency is toward the use of large rivets, and, conse- 
quently, large pitches, the excess tenacity applicable in practi- 
cal cases becomes small, and would be better disregarded in 
the design of most riveted joints. In cases where the holes 
are drilled, therefore, we should use for tensile strength per 
square inch of the plate along the line of rivet-holes, the tensile 
strength per square inch of the plate itself. 



COMPRESSION. 637 



The better and more ductile the plate the less is the 
injury done by punching; but, while more or less punching is 
done, the better class of work is drilled. A study of the results 
in the cases of punched plates will show approximately what 
allowance to make for the weakening due to punching different 
qualities of plate. 

Shearing. — A study of the results of the government tests 
show that it is fair to assume the shearing-strength of the 
wrought-iron rivets used, to be about 38000 pounds per square 
inch, which is about two thirds of the tensile strength of the 
same rivet metal. 

For steel rivets, of the kinds now prescribed in most spec- 
ifications, the shearing-strength appears to be about 45000 
pounds per square inch. 

Compression. — To determine what we should estimate as 
the ultimate compression on the bearing-surface is a more 
difficult problem ; for if a joint fails in consequence of too 
great compression on the bearing-surface the cause of the 
failure does not exhibit itself directly, but in some indirect 
manner — probably by decreasing the resisting properties of 
either the plate or the rivets, and hence by causing either the 
joint to break by tearing the plate or by shearing either the 
rivets or the plate in front of the rivets, but at a lower load 
than that at which it would have broken had the compression 
not been excessive ; and hence when such breakage occurs it is 
difficult to say whether it is due to excessive compression re- 
ducing the tensile or the shearing strength, or whether its full 
tensile or shearing strength was really reached. 

Observe, moreover, that in the tables of Government tests 
the heavy numbers in the column marked " Compression on 
the bearing-surface of the rivets " indicate that the plate broke 
out in front of the rivets, which might be due to excessive 
compression or to a deficiency of lap. 

While more experiments are needed, it would seem proba- 
ble that we might deduce some conclusions, at least, of a gen- 
eral nature, in regard to the ultimate compression by a study 



638 APPLIED MECHANICS. 

of the relations existing between the compression per square 
inch on the bearing-surface at fracture and the efficiency of 
the joint as shown by the Government tests. 

For this purpose the following diagrams (see pages 631 and 
632) have been plotted, with the efficiencies as abscissae and 
the compression per square inch on the bearing-surface at 
fracture as ordinates. If similar diagrams were plotted with 
the efficiencies as abscissae and the ratio of the compression per 
square inch on the bearing-surface at fracture to the tensile 
strength of the plate as ordinates the character of the diagrams 
would be substantially the same, as the plates used in the tests 
were all of mild steel of approximately the same quality, and 
hence the difference in tensile strength of different samples 
was not great. 

A study of these diagrams shows that in the case of the 
i-inch plates experiments were made with compressions up to 
about 158,000 pounds per square inch, but that the highest 
compression reached with any other thickness of plate was 
about 120,000 pounds per square inch. 

Inasmuch as Kennedy advises the use of 96,000 pounds per 
square inch, and as this is higher than the values that have 
been customarily advocated, it would hardly seem wise to 
adopt a much higher value unless the tests furnish us sufficient 
evidence for such a procedure. Considering the facts stated 
above, and also the fact that in the cases of the double-riveted 
joints some of the highest compressions were accompanied by 
a decrease in efficiency, it would seem best to limit our esti- 
mate of the ultimate compression on the bearing-surface to 
from 90,000 to 100,000 pounds per square inch until we have 
further light on the subject derived from experiment ; and it is 
not at all improbable that when we do obtain further light we 
may find ourselves warranted in using a somewhat higher 
value. 

The reasoning which leads to the above conclusion is, of 
course, based on evidence which is not conclusive, because of 
the lack of tests with higher compressions on the bearing sur- 



COMPRESSION 



638* 



co\rpf?£ss/ov 0<vj3£yttT/#Gst//rS4C£ or at vers 




6$U 



APPLIED MECHANICS. 



0OUBLE-P/VETE0 Bt/TT-<JO/A/T$ 
W/TH TWO COy£/?//VG ST/F/PS. 
■£" STEEL PLATE. $' 'STEEL PLATE. 




$0000, 



50000 



70 60 

face, with plates thicker than one quarter of an inch. On the 
other hand, the quarter-inch plates show higher efficiencies 
with compressions above iooooo pounds than they do with 
compressions of iooooo pounds or less, and the author knows 
of tests upon riveted joints in T 7 ¥ -inch plates which tend to 
show that, with good wrought-iron rivets, it would be perfectly- 
safe to use a considerably larger number for compression on 
the bearing-surface, in designing riveted joints — at least 
Iioooo pounds per square inch, and probably more. 



COMPRESSION. 



638' 



It will be observed that no reference has been made to the 
friction, and it is safer to leave this out of account, as the tests 
show thac slipping takes place at all loads, and as there is no 
friction at the time of fracture. 

By far the greater part of the tests at Watertown Arsenal 
were made with wrought-iron rivets in mild-steel plates, this 
being, at the time, the most usual practice, although steel 
rivets were sometimes used. At the present time, notwith- 
standing the fact that steel long ago superseded wrought- 
iron for boiler-plate, and that it has, to-day, superseded 
wrought-iron for structural shapes, as I beams, channel-bars, 
angles, etc., and that the use of steel rivets has become very 
extensive, nevertheless a great many still adhere to the use 
of wrought-iron rivets, and feel more confidence in them than 
they do in steel rivets. Whereas the use of wrought-iron 
rivets had been practically universal, the qualifications for a 
good wrought-iron rivet metal became pretty well known, and 
while sometimes specifications were drawn up giving the 
requirements of the rivet metal for tensile strength, ductility, 
etc., which of course would vary more or less, nevertheless 
the variations would not be large. A study of the Watertown 
tests shows that the wrought-iron rivet metal used in those 
tests had a tensile strength of from about 52000 to about 
59000 pounds per square inch, with a percentage contraction 
of area at fracture of from about 30 to about 45. With this 
metal the shearing strength per square inch seems to be about 
f of the tensile strength per square inch. Of course other 
tests are necessary to show whether the metal can be properly 
worked, and whether it is red-short or not, such as that the 
metal should bend double, whether cold or hot, without cracks, 
and that cracks should not develop when the shank is ham- 
mered down, cold or hot, to a length considerably less than 
the diameter. 

When steel rivets were first used, the steel employed was 



63 8 d APPLIED MECHANICS. 

not an extremely soft steel, as shown by the few cases of steel 
rivets included in the Watertown Arsenal tests already quoted, 
where the shearing -strength per square inch varied from about 
50000 pounds per square inch up to as high a figure as 65000 
pounds per square inch; and by Kennedy's tests, where he ap- 
parently fixes on from about 49000 to about 54000 pounds per 
square inch as the shearing-strength of steel rivets. 

Now it would seem that metal with these shearing-strengths 
would have a tensile strength per square inch which would not 
warrant us in classifying it as very soft steel. 

On the other hand, it is evident that brittleness should not 
in any way be tolerated in rivet metal, and hence it would seem 
that at least soft steel should be used for rivets. 

The specifications proposed by the American Society lor 
Testing Materials prescribe for tensile strength per square inch 
of • steel for structural rivets from 50000 to 60000 pounds per 
square inch, and for boiler-rivets from 45000 to 55000 pounds. 

While the number of tests that have been made upon joints 
constructed with steel rivets is not large, the shearing-strength 
of such steel rivets as are in use to-day is not very far from 45000 
pounds per square inch, as a rule. 

The number of tests of joints constructed with steel rivets is 
not sufficiently large to warrant drawing from them definite 
conclusions regarding the ultimate compression on the bearing 
surface in such joints. Meanwhile, it would be advisable to use 
for it the same values as are suitable in the case of joints made 
with steel plates and wrought-iron rivets. 

The following table contains the joints tested at Watertown 
Arsenal, which were made with steel plate and wrought-iron 
rivets, and in which the plate broke out in front of the rivet. It 
is evident that only four of them, viz., 4915, 4916, 4917, 4918, 
failed in consequence of excessive compression on the bearing 
surface, and that the breaking out of the plate in the other cases 

was due to insufficiency of lap. The calculated -3 was obtained 



WIRE AND WIRE ROPE. 



639 



by the method described on page 554, assuming /* = 55000, and 
j 8 = 38000, and } c = 96000. 






718 

719 
4947 
4948 
4949 

767 
1442 
1443 
49 

4916 
49 

4918 
4985 
4987 
4991 

'298 



5 

51 22 



Kind of 
Joint. 



Single lap 



Single butt 



Reinforced 
lap 



Double lap 



Iron 



£ O 

Q 
Ins. 



OhG 






H 
Ins. 



UJ ft 



Ins. 



Ins. 



1. 25 
2.00 
2.00 
2.00 
1.25 
2.00 
2.00 
2 .00 



.25 
•75 

• 15 
.00 
. 10 
. 12 

• 50 

• 50 



•a > 

I" 

Ins. 



25 
2.00 
2 .00 
2 .00 
2 .00 
2 .00 
2 .00 
1 .00 
125 
i-75 
1 .12 






Ifll 

3°"~ 

Lbs. 



795io 

80200 
112970 
1 1 47 60 
120180 

95210 
107610 
108830 
151780 
142570 
15815° 
152110 

83950 

937io 
119500 

67300 
68040 

86180 
86450 



§■3 

£^ 

O 



Tore and 

sheared 

Tore 



Tore 
Tore 



Sheared 
rivets 



1. 18 



1 .60 
1 .60 
1 .60 
1.67 
1.50 
1.50 
2 . 29 
2. 29 
2. 29 
2. 29 
1 .00 
1-25 
1-75 

1. 19 



T.50 
1.50 



i.55 

i.55 
1.75 
1.75 
1.77 
1.6 

T . 69 

1.70 
1.93 
I.89 
I.96 
1.93 
1-57 
1.63 

1-77 
1.66 

1.67 
1.56 



§ 234. Wire and Wire Rope. — It is well known that the 
process of making wire by cold drawing greatly increases the 
strength of the metal. Annealing, on the other hand, decreases 
the strength, and increases the ductility. It is not the purpose 
of this article to discuss the various qualities of wire required and 
used for different purposes. Hence, inasmuch as results of tests 
of wrought-iron, and of steel wire, have already been given, there 
will be given here only a few tests of hard-drawn, of semi-hard- 
drawn, and of soft copper wire. 

Wire rope. — Wire rope is used for a great many purpose?, as 
in suspension bridges, in hoisting, in haulage, in the transmission 
of power, etc. 

While flat wire rope is used for some purposes, and while 
wire rope made of parallel wires is used in large suspension 
bridges, the greater part is made by twisting a number of wire 



640 



APPLIED MECHANICS. 



HARD-DRAWN COPPER WIRE. 



SOFT COPPER WIRE. 



Diameter. 
Inches. 


Tensile 
Strength 

per 

Sq. In. 

Lbs. 


Elastic 

Limit per 

Sq. In. 

Lbs. 


Contrac- 
tion of 
Area. 

Per Cent. 


0. 166 
0.138 
0-I35 
O.I34 
0.105 
0.105 
O.105 
0. 106 
O.I06 
O.086 
O.086 
O.085 
O.083 
O.083 
O.083 


53050 
60350 
56300 

5 IQ 5o 

61800 
57IOO 
58900 
60300 
5950O 
58170 
58620 
61510 
66536 
65060 
66536 


37100 
22800 
28150 
2,7140 
410CO 
35000 
34CCO 
36000 
34000 
27870 
29310 
21390 
29630 

37334 
29630 


51 
49 
33 
42 
39 



Diameter. 
Inches. 


Tensile 
Strength 

per 
Sq. In. 

Lbs. 


Elastic 

Limit per 

Sq. In. 

Lbs. 


Contrac- 
tion of 
Area. 

Per Cent. 


O.163 
0. 162 
O. 162 
O.083 
0.083 
O.081 
O.080 
O.080 


35730 
35770 
36640 

2 9500 
29500 
332 00 

33IOO 
33080 


13740 
13760 
12990 
10500 

IJ2CO 


7o 
45 
7o 



SEMI-HARD-DRAWN COPPER WIRE. 



0. 106 


44300 


30000 


60 


0. 106 


45100 


290C0 


65 


0. 106 


455°° 


29000 


55 


0.106 


45100 


31000 


67 


0. 106 


44900 


30000 


64 



strands around a central core, which niay be of tarred hemp, or 
which may be, itself, a wire strand, the wire strands being made 
of wires twisted together. 

In the case of a wire core, the strength of the rope is a little 
greater, but the resistance to wearing is less. 

The most usual number of strands is six, each strand contain- 
ing seven, eighteen, or nineteen wires, though other numbers of 
wires are sometimes used. 

The strength that can be realized in practice is always less 
than the strength of the rope, and is determined by the method of 
holding the ends, as the junction point of the rope and the holder 
is the weakest point. 

The usual methods of holding the ends are as follows: splic- 
ing, as in the case of the transmission of power, passing the rope 
around a pulley, or around a thimble, fastening it in a socket, or 
in a clamp. 

The diameter of the drum or sheave around which a rope 



WIRE AND WIRE ROPE. 



64I 



passes, should not be so small as to cause too much stress to be 
exerted upon some of the wires, in consequence of the bending- 
moment introduced by the curvature. 

Inasmuch as it may be a matter of convenience to have here 
some tables giving the strength of rope as claimed by some makers, 
there will follow here two tables of the strength of different sizes, 
as given by the Roebling Company for their rope. 

The following explanations are given by the Roebling Com- 
pany, about the quality of the metal used : 

Iron, open-hearth steel, crucible steel, and plough steel pos- 
sess qualities which cover almost every demand upon the material 
of a wire rope. Copper, bronze, etc., are, however, used for a 
few special purposes. 

The strength of iron wire ranges from 45000 to 100000 pounds 
per square inch; open-hearth steel, from 50000 to 130000 pounds 

SEVEN-WIRE ROPE. 
Composed of 6 Strands and a Hemp Center, 7 Wires to the Strand. 











Approximate Breaking-strain in Tons of 






Approxi- 




2000 Lbs. 










Trade No. 


Diameter 
in Inches. 


mate 
Circum- 
ference 


Weight _ 

per Foot in 

Pounds. 


Transmission or 
Haulage Rope. 


Extra 


Plough 
Steel. 






in Inches. 








Strong 










Swedish 


Cast 


Cast Steel. 










Iron. 


Steel. 






11 


*i 


4| 


3-55 


34 


68 


79 


91 


12 


if 


4i 


3.00 


29 


58 


68 


78 


\3 


ii 


4 


2.45 


24 


48 


56 


64 


14 


ii 


2>h 


2 .00 


20 


40 


46 


53 


IS 


1 


3 


158 


16 


3 2 


37 


42 


16 


7 
t 


2! 


1 . 20 


12 


24 


28 


32 


17 


3 
4 


2i 


0.89 


9-3 


18.6 


21 


24 


18 


11 


2i 


o.75 


7-9 


15-8 


18.4 


21 


19 


I 


2 


0.62 


6.6 


13.2 


15 


1 


17 


20 


& 


if 


0.50 


5-3 


10.6 


12 


3 


14 


21 


h 


I* 


o.39 


4.2 


8.4 


9 


70 


II 


22 


& 


ii 


0.30 


3-3 


6.6 


7 


50 


8-55 


23 


t 


ii 


0.22 


2.4 


4-8 


5 


58 


6-35 


24 


& 


1 


°-i5 


i-7 


3-4 


3 


88 


4-35 


1 2 5 


■h 


1 


0.125 


1.4 


2.8 


3 


22 


3-6 5 



642 



APPLIED MECHANICS. 



NINETEEN-WIRE ROPE. 
Composed of 6 Strands and a Hemp Center, 19 Wires to a Strand. 











Approximate Breaking-strain in Tons of 










2000 Lbs. 








Approxi- 


Weight. 








Trade No. 


Diameter 
in Inches. 


mate 
Circum- 
ference 
in Inches. 


per Foot in 
Pounds. 


Standard Hoisting 
Rope. 


Extra 
Strong 


Plough 
Steel. 














Swedish 


Cast 


Cast Steel. 












Iron. 


Steel. 






— 


2| 


8f 


n-95 


— 


— 


— 


305 


— 


4 


li 


9-85 


— 


— 


■ — 


254 


1 


a£ 


7* 


8.00 


78 


156 


182 


208 


2 


2 


6i- 


6.30 


62 


124 


144 


105 


3 


If 


5* 


4.85 


48 


96 


112 


128 


4 


If 


5 


4-15 


42 


84 


97 


ill 


5 


ii 


4f 


3-55 


36 


72 


84 


96 


5* 


if 


4i 


3.00 


31 


62 


72 


82 


6 


ii 


4 


2-45 


25 


5° 


58 


67 


7 


ii 


3* 


2 .00 


21 


42 


49 


56 


8 


1 


3 


1.58 


17 


34 


39 


44 


9 


I 


2f 


1 .20 


13 


26 


30 


34 


10 


3 

4 


ai 


0.89 


9-7 


19.4 


22 


25 


ioi 


f 


2 


0.62 


6 


8 


13 


6 


15-8 


18 


io| 


& 


if 


0.50 


5 


5 


11 





12.7 


14.5 


iof 


1 


i| 


o-39 


4 


4 


8 


8 


10. 1 


11. 4 


10a 


A 


if 


0.30 


3 


4 


6 


8 


7-8 


8.85 


10b 


f 


ii 


0. 22 


2 


5 


5 





5-78 


6-55 


IOC 


ft 


1 


0.15 


1 


7 


3 


4 


405 


4-5° 


lod 


1 


f 


0. 10 


1 


2 


2 


4 


2. 70 


3.00 



per square inch; crucible steel from 130000 to 190000 pounds 
per square inch; and plough steel from 190000 to 350000 pounds 
per square inch. Plough steel wire is made from a high grade of 
crucible cast-steel. 

§ 235. Other Metals and Alloys. — Copper is, next to iron 
and steel, the metal most used in construction, sometimes in the 
pure state, especially in the form of sheets or wire, but 
more frequently alloyed with tin or zinc; those metals where 
the tin predominates over the zinc being called bronze, and 
those where zinc predominates over tin, brass. Copper in the 
pure state was used not long ago for the fire-box plates of loco- 



IRON AND STEEL WIRE. 643 

motive and other steam-boilers, "as it was believed to stand better 
the great strains due to the changes of temperature that come 
upon these plates, than iron or steel ; but now steel or iron has 
almost entirely superseded it for this purpose, except in some 
cases where the feed-water is very impure, and where the 
impurities are such as corrode iron. 

The alloys of copper, tin, and zinc which are used most 
where strength and toughness are needed, are those where the 
tin predominates over the zinc ; and the composition, mode of 
manufacture, and resisting properties of these metals, together 
with the effect of other ingredients, as phosphorus, have been 
very extensively investigated with reference to their use as a 
material for making guns, instead of cast-iron. 

Accounts of tests made on these alloys will be found as 
follows : — 

Major Wade : Ordnance Report, 1856. 

T. J. Rodman : Experiments on Metals for Cannon. 

Executive Document No. 23, 46th Congress, 2d session. 

Materials of Engineering : Thurston. 

No attempt will be made to give a complete account of the 
results of these tests ; but a table will be given on page 639 for 
convenience of use, showing rough average values of the resist- 
ing powers of some metals and alloys other than iron. 

§236. Timber. — However extensively iron and steel may 
have superseded timber in construction, nevertheless, there are 
many cases in which iron is entirely unsuitable, and where 
timber is the only material that will answer the purpose ; and 
in many cases where either can be used, timber is much the 
cheaper. Hence it follows that the use of timber in construc- 
tion is even now, and as it seems always will be, a very impor- 
tant item. 

Another advantage possessed by timber is, that, on yielding, 
it gives more warning than iron, thus affording an opportunity 
to foresee and to prevent accident. 

If we make a section across any of the exogenous trees, as 



■6 4 4 



APPLIED MECHANICS. 



Specific 
Gravity. 



Tensile 
Strength 
per Sq. In. 



Modulus 

of 
Elasticity. 



Brass cast 

Brass wire 

Bronze unwrought : 

84.29 copper +15.71 tin (gun metal) 

82.81 " + 17.19 " 

81.10 " + 18.90 " 

78.97 " + 21.03 " (brasses) . . . 

34.92 " + 65.08 " (small bells) . 

15.17 " +84.83 " (speculummetal) 

Tin 

Zinc 

Copper cast 

Copper bolts 

Copper wire 

Gold cast 

Silver cast 

Platinum wire 

Lead cast 



8.396 



8.561 
8.462 

8-459 
8.728 
8.056 

7447 
7.291 
6.861 
8.712 



19.258 
10.476 
22.069 
"•35 2 



18000 
49000 

36060 

34048 

39648 

30464 

3^6 

6944 

5600 

7500 

24138 

33000 

60000 

20000 

40000 

56000 

1800 



9170000 
[4230000 



17000000 



the oak, pine, etc., we shall find a series of concentric layers ; 
these layers being called annual rings, because one is generally 
deposited every year. 

Radiating from the heart outwards will be found a series of 
radial layers, these being known as the medullary rays. 

Of the annual rings, the outer ones are softer and lighter in 
color than the inner ones ; the former forming the sap-wood, and 
the latter the heart-wood. When the log dries, and thus tends 
to contract, it will be found that scarcely any contraction takes 
place in the medullary rays ; but it must take place along 
the line of least resistance, viz., along the annual rings, thus 
causing radiating cracks, and drawing the rays nearer together 
on the side away from the crack. This action is exhibited in 
Fig. 241, where a log is shown with two saw-cuts at right 
angles to each other ; when this log be ^mes dry, the four 



STRENGTH OF TIMBER. 



64 




Fig. 241 



right angles all becoming acute 
through the shrinkage of the 
rings. 

If the log be cut into planks by 
parallel saw-cuts, the planks will, 
after drying, assume the forms 
shown in Fig. 242, as is pointed 
out in Anderson's " Strength of 
Materials," from which these two 
cuts are taken. 

This internal construction of a 
plank has an important influence 
upon the side which should be uppermost when it is used for 

flooring ; for, if the heart side is up- 
permost, there will be a liability to 
having layers peel off as the wood 
dries : indeed, boards for flooring- 
should be so cut as to have the an- 
nual rings at right angles to the 
side of the plank. Before discuss- 
ing any other considerations which 
affect the adaptability of timber to 
use in construction, we will con- 
sider the question of its strength. 
§ 237. Strength of Timber. — In this regard we must 
observe, that, whereas the strength and elasticity and other 
properties of iron and steel vary greatly with its chemical com- 
position and the treatment it has received during its manufac- 
ture, the strength, etc., of timber is much more variable, being 
seriously affected by the soil, climate, and other accidents of its 
growth, its seasoning, and other circumstances ; and that over 
many of these things we have no control : hence we must not 
expect to find that all timber that goes by one name has the 
same strength, and we shall find a much greater variation and 




Fig. 242. 



646 APPLIED MECHANICS. 

irregularity in timber than in iron. The experiments that have 
been made on strength and elasticity of timber may be divided 
into the following classes : — 

i°. Those of the older experimenters, except those made 
on full-size columns by P. S. Girard, and published in 1798. 
A fair representation of the results obtained by them, all of 
which were deduced from experiments on small pieces, is to 
be found in the tables given in Professor Rankine's books, 
"Applied Mechanics," " Civil Engineering," and "Machinery 
and Millwork." 

2 . Tests made by modern experimenters on small pieces. 
Such tests have been made by — 

(a) Trautwine : Engineers' Pocket-Book. 

(b) Hatfield : Transverse Strains. 

(c) Laslett : Timber and Timber Trees. 

(d) Thurston : Materials of Construction. 

(e) A series of tests on small samples of a great variety of American 

woods, made for the Census Department, and recorded in 
Executive Document No. 5, 48th Congress, 1st session. 
Timber Physics, Division of Forestry, U. S. Department of Agri- 
culture. For a fairly complete bibliography of tests of tim- 
ber see a paper by G. Lanza, Trans. Am. Soc. C. E., 1905. 

3 . Tests made by Capt. T. J. Rodman, U.S.A., the results 
of which are given in the "Ordnance Manual." 

4 . All tests that have been made on full-size pieces. 

In regard to tests on small pieces, such as have commonlv 
been used for testing, it is to be observed, that, while a great 
deal of interesting information may be derived from such tests 
as to some of the properties of the timber tested, nevertheless, 
such specimens do not furnish us with results which it is safe 
to use in practical cases where full-size pieces are used. Inas- 
much as these small pieces are necessarily much more perfect 
(otherwise they would not be considered fit for testing), having 
less defects, such as knots, shakes, etc., than the full-size pieces. 



STRENGTH OF TIMBER. 



647 



they have also a far greater homogeneity. They also season 
much more quickly and uniformly than full-size pieces. In 
making this statement, I am only urging the importance of 
adopting in this experimental work the same principle that the 
physicist recognizes in all his work ; viz., that he must not 
apply the results to cases where the conditions are essentially 
different from those he has tested. 

Moreover, it will be seen in what follows, that, whenever 
full-size pieces have been tested, they have fallen far short of 
the strength that has been attributed to them when the basis 
in computing their strength has been tests on small pieces ; 
and, moreover, the irregularities do not bear the same propor- 
tion in all cases, but need to be taken account of. 

The results of the first class of experiments named in the 
following table are taken from Rankine's " Applied Mechanics;" 
and, inasmuch as the table contains also the strengths of some 
other organic fibres, it will be inserted in full. The student 
may compare these constants with those that will be given 
later. 



Kind of Material. 


Tenacity 
or Resist- 
ance to 
Tearing. 


Modulus of 
Tensile 

Elasticity. 


Resist- 
ance to 
Crush- 
ing. 


Modulus 

of 
Rupture. 


Resist- 
ance to 
Shearing 
along 
Grain. 


Modulus 

of 

Shearing 

Elasticity 

along the 

Grain. 


Ash ...... . 

Bamboo 

Beech 

Birch 

Blue gum 

Box 

Bullet-tree .... 
Cedar of Lebanon . . 


I70OO 

63OO 

1 1 500 

1500O 

20000 
I I4OO 


ICOOOOO 

I35OOOO 
1645OOO 

4860OO 


9000 

9360 
64OO 
8800 
IO30O 
14000 
5860 


( I2000 
( 1400O 

( 9000 
( 12000 
1 1 700 
{ IbOOO 
( 2O00O 

( I59OO 

\ 16000 

7400 


1400 


76000 



648 



APPLIED MECHANICS. 



Kind of Material. 



Tenacity 
or Resist- 
ance to 
Tearing. 



Modulus of 

Tensile 
Elasticity. 



Resist- 
ance to 
Crush- 
ing. 



Modulus 

of 
Rupture. 



Resist- 
ance to 
Shearing 
along 
Grain. 



Modulus 

of 
Shearing 
Elasticity 
along the 



Chestnut 



Cowrie 

Ebony 



Elm 



Fir, Red pine . . 
" Yellow pine (Am 
" Spruce . . 

" Larch . . 

Hoxen yarn . . 
Hazel .... 



Hempen rope 

Ox-hide, undressed 
Hornbeam . . 
Lancewood . . 
Ox-leather . . 
Lignum-vitae . . 
Locust .... 



Mahogany . . . 

Maple .... 
Oak, British . . 
" Dantzic 

" European . 

" American red 



1 0000 

to 
13000 



14000 

12000 

to 
14000 



12400 



1 0000 

to 
19800 
10250 



1 140000 



700000 

to 
1840000 
1460000 

to 
1900000 



900000 

to 
I 360000 



24300 



1255000 



I 200000 
1750000 

2150000 



19000 
10300 

5375 

to 

6200 

5400 



5570 



9900 
8200 

1 0000 
7700 

6000 



[0660 



1 1000 

27000 

6000 

to 

9700 

7100 

9540 



j 9900 
\ 12300 

\ 5000 

\ I OOOO 



7600 
1 1 500 



1 0000 

13600 

8700 



[0600 



1400 



600 



970 

1700 



^2300 



76000 



62000 
[ 16000 



82000 



STRENGTH OF TIMBER. 



649 



Kind of Material. 


Tenacity 
or Resist- 
ance to 
Tearing. 


Modulus of 

Tensile 
Elasticity. 


Resist- 
ance to 
Crush- 
ing. 


Modulus 

of 
Rupture. 


Resist- 
ance to 
Shearing 
along 
Grain. 


Modulus 

of 

Shearing 

Elasticity 

along the 

Grain. 


Silk fibre . . . , 


52000 
I300O 

150OO 

2IOOO 
77OO 

8000 


I30OOOO 
IO400OO 

240OOOO 

2300OOO 


12000 


9600 
( I2O0O 
I I900O 

14980 
660O 


\ l 


- 


Sycamore . . . 

Teak, Indian 

" African . 
Whalebone . . 
Willow. . . . 







In regard to the tests of the second class, a few comments 
are in order : — 

i°. These experiments, like those of the first class, were all 
made upon small pieces ; and the results are correspondingly 
high. 

The usual size of the specimens for crushing being one or 
two square inches in section, and of those for transverse 
strength being about two inches square in section and four or 
five feet span, those for tension had even a much smaller sec- 
tion than those for compression ; as it is necessary, in order to 
hold the wood in the machine* to give it very large shoulders. 

The only exception to this is the tests of Sir Thomas Las- 
lett, an account of which is given in his " Timber and Timber 
Trees," and also in D. K. Clark's " Rules and Tables." In these 
tests he gives very much lower tensile strengths than those 
given above ; and he states that his specimens were three inches 
square, but does not say how he managed to hold them in such 
a way as to subject them to a direct tensile stress. His results 
for crushing and transverse strength are about as great as 



^5° APPLIED MECHANICS. 

those given in Rankine's tables, and as were obtained by the 
other experimenters on small pieces, as his specimens were of 
about the same dimensions as those used by the others. The 
figures obtained by these experimenters will only be given inci- 
dentally, as — 

(a) They are very similar to those given in Rankine's table. 

(b) They are not suitable for practical use on the large 
scale. 

(c) While they have been used, it has only been done by 
employing a very large factor of safety for timber. 

The series of tests made for the Census Department, and 
recorded in Executive Document No. 5, 48th Congress, first 
session, form a very interesting series of experiments upon 
small specimens of an exceedingly large number of American 
woods. In order to have working figures, we should need to 
test large pieces of the same ; as the proportion between the 
strengths of the different kinds would be liable to be different 
in the latter case. 

The work done by the Division of Forestry of the U. S. 
Dept. of Agriculture before 1898 was mostly of this class, but 
little having been done with full-size pieces, and that with 
imperfect apparatus. 

The only record of Rodman's experiments available is a 
table of results in the " Ordnance Manual." These are lower, 
as a rule, than those obtained by the experimenters of the first 
or second class. This is to be accounted for by the fact that, 
while he did not experiment on full-size pieces, he used much 
larger pieces than those heretofore employed ; his specimens 
for transverse strength, many of which are still stored at the 
Watertown Arsenal, being '5f inches deep, 2$ inches thick, and 
5 feet span. 

The fourth class of tests are those which furnish reliable 
data for use in construction ; and we will proceed to a consid- 
eration of these, taking up (i°) tension, (2 ) compression, (3 ) 
transverse strength, and (4 ) shearing along the grain. 



TENSION. 65 1 



TENSION. 

In all cases where the attempt has been made to experiment 
upon the tensile strength of timber, a great deal of difficulty 
has been encountered in regard to the manner of holding the 
specimens. In all cases it has been found necessary to pro- 
vide them with shoulders, each shoulder being five or six times 
as long as the part of the specimen to be tested, and to bring 
upon these shoulders a powerful lateral pressure, to prevent 
the specimen from giving way by shearing along the grain, and 
pulling out from the shoulder, instead of tearing apart. 

The specimens tested have generally had a sectional area 
less than one square inch, and it seems almost impossible to 
provide the means of holding larger specimens. This being 
the case, it is plain, that, whenever timber is used as a tie-bar 
in construction (except in exceedingly rare and out-of-the- 
way cases), it will give way by some means other than direct 
tension ; i.e., either by the pulling-out of the bolts or fastenings, 
and the consequent shearing of the timber, or else by bending 
if there is a transverse stress upon the piece ; and, this being 
the case, these other resistances should be computed, instead 
of the direct tension. Hence, while the direct tensile strength 
of timber may be an interesting subject of experiment, it can 
serve hardly any purpose in construction ; and the conclusion 
follows, that the resistances of timber to breaking we may 
expect to meet in practice are its crushing, transverse, and 
shearing strength. Indeed, the use of timber for a tie-bar 
should be avoided whenever it is possible to do so ; and, when 
it is used, the calculations for its strength should be based 
upon the pulling-out of the fastenings, the shearing or splitting 
of the wood, etc., and not on the tensile resistance of the solid 
piece. 

Moreover, when a wooden tie-bar is subjected not merely 
to direct tension, but also to a bending-moment, whether the 
latter is caused by a transverse load, or by an eccentric pull, as 
it generally is in the case of timber joints, we must compute 



652 APPLIED MECHANICS. 

the greatest tension per square inch at the outside fibre due to 
the bending,. and to that add the direct tension per square inch: 
and this sum must be less than the modulus of rupture if the 
piece is not to give way; i.e., the modulus of rupture and not 
the ultimate tensile strength per square inch must be our criterion 
of breaking in such a case, the working-strength per square inch 
being the modulus of rupture divided by a suitable factor of safety. 

COMPRESSIVE STRENGTH. 

Tests of the compressive strength of full-size wooden columns, 
with the exception of one set of tests, date from about 1880. 

TESTS OF FULL-SIZE COLUMNS. 

The following are references to tests of full-size timber columns : 

i°. Trautwine, in his "Handbook," speaks of some tests of 
wooden pillars 20 feet long and 13 inches square, made by David 
Kirkaldy, which, as he says, gave results agreeing with Mr. C. 
Shaler Smith's rule. 

2 . A series of tests made at the Watertown Arsenal for the 
Boston Manufacturers' Mutual Fire Insurance Company, under 
the direction of the author. 

3 : Eleven tests of old spruce pillars made at the Watertown 
Arsenal, for the Jackson Company, under the direction of Mr. 
J. R. Freeman, and reported in the Journal of the Assoc. Eng. 
Societies for November, i< 

on the government testing-machine. 

5 . Tests made in the Laboratory of Applied Mechanics of 
the Massachusetts Institute of Technology. 

6°. A series of tests of full-size columns of oak and fir, made 
by P. S. Girard in 1798. 

In regard to the first, no details or results are given: hence 
nothing will be said about them. 

In regard to the second, a summary only will be presented 
here. 



TESTS OF YELLOW-PINS POSTS AND BLOCKS. 



655 



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TESTS OF OLD AND SEASONED WHITE-OAK POSTS. 655 



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656 



APPLIED MECHANICS. 



In all the experiments enumerated in the tables given 
above, the columns gave way by direct crushing, and hence 
the strength of columns of these ratios of length to diameter 
can properly be found by multiplying the crushing-strength per 
square inch of the wood by the area of the section in square 
inches. 

This conclusion is deduced from the fact that the deflections 
were measured in every case, and found to be so small as not to 
exert any appreciable effect. 

In regard to other tests of this same set, there were eight 
tests made, in addition to those already enumerated; and in 
five the loads were off centre. A summary of the results is 
appended, together with a comparison of their actual strength 
with that which would be computed on the basis of 4400 per 
square inch for yellow pine, and 3000 for oak. The first three 
tests were made on yellow-pine columns, and the last two on 
oak. 





Weight, 
in lbs. 


Length, in 
feet and 
inches. 


Diameter 

of 
Column. 


Diam- 
eter of 
Core. 


Sectional 
Area, in 
square 
inches. 


Eccen- 
tricity, 

in 
inches. 


Ultimate 
Strength. 


Computed 
Ultimate 
Strength. 






ft. in. 














2, 2d series 


320 


II II.27 


9.92 


i-S3 


7545 


2-33 


2650OO 


331980 


5, 3d series 


298 


12 6.8 


( 8.30) 

\ X ( 
( 7.60) 


- 


63.1 


2.07 


2400OO 


277640 


1, 3d series 


386 


12 9.3 


I 8.92) 


- 


76.04 


2.25 


280000 


334576 


1, 2d series 


451 


11 11.4 


10.95 


1.80 


92.16 


275 


170000 


276480 


3, 2d series 


236 


11 11. 2 


8.2 


i-55 


50.92 


1. 91 


I OOOOO 


152760 



These results exhibit a great falling-ofr of strength due to 
the eccentricity of the load ; and emphasizes the importance 
of taking into account eccentric loading in our calculations in 
a manner similar to that already mentioned on pages 370, 371, 
and 448. 



STRENGTH OF TIMBER. 657 

The remaining experiments were : (i°) Two tests of white- 
wood columns, average strength 3000 pounds per sq. in., and 
very brittle. (2 ) One yellow-pine square column (sectional area 
68.8 sq. in., length 12' 6 // .85) with one end resting against a 
thick yellow-pine bolster. 

The maximum load was 120,000 lbs. = 1744 lbs. per sq. 
in., the post beginning to split due to eccentricity of bearing 
caused by uneven yielding of the bolster. The bolster was 
then removed, the post cut off i-J- in. at the end and tested 
without the bolster. Ultimate strength 375,000 lbs. = 5451 
lbs. per sq. in. 

The table of results of the tests on old and seasoned oak 
columns were made upon columns that had been in use for a 
number of years in different mills, from which they were re- 
moved, and replaced by new ones. Ten of them had been in 
use about twenty-five years, and the remainder for shorter 
periods. An inspection of this table will, I think, convince the 
reader that it would not be safe to calculate upon a higher 
breaking-strength per square inch in these than in the green 
ones. 

TESTS FOR THE JACKSON COMPANY. 

Eleven tests of old spruce pillars, which had been in use 
in a cotton-mill of the Jackson Company, were tested on the 
government machine at Watertown, under the direction of 
Mr. J. R. Freeman. The manner of making them was as fol- 
lows : 

In the first two the ends were brought to an even bear- 
ing. 

In the third the ends came to an even bearing under a load 
of 60000 pounds. 

In the fourth, fifth, ninth, tenth, and eleventh, the cap, 
and also the base-plate, were planed off on the back to a slope 
of 1 in 24, and placed with their inclinations opposite. 

In the eighth they had their inclinations the same way one 
as the other. 



658 



APPLIED MECHANICS. 



In the sixth and seventh the base-plate was not used, the 
larger end of the post having a full bearing on the platform of 
the machine. 

The results are given in the following table : 



... 












" -^ 




Length 

in Feet and 

Inches. 


Diameter 

at 

Small End, 

Inches. 


Diameter 

at 

Large End, 

Inches. 


Area at 

Small End, 

Sq. In. 


Ultimate 

Strength, 

Lbs. 


Ultimate 

Strength per 

Sq. In., 

Lbs. 




ft. in. 












I 


10 4-75 


5.82 


'7-78 


31.87 


T 42000 


4088 


2 


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5 


85 


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27.15 


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6 


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61 . 


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81 


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8 


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90 


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9 


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77 


40.72 


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10 


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5 


73 


7 


78 




125000 




" 


TO 4.38 


5 


74 


7 


81 




60000 





All but the first three of the tests were made with inclined 
bearings of one kind or another, hence the ultimate strength 
per square inch is only given here for the first three ; which, as 
Mr. Freeman says, were of " well-seasoned spruce, of excellent 
quality." Hence the average crushing-strength of spruce is 
doubtless considerably lower than the average of these three. 



TESTS MADE ON THE GOVERNMENT MACHINE. 

In Executive Document 12, 47th Congress, first session, 
will be found a series of tests of white and yellow pine posts 
made at the Watertown Arsenal; and these tests probably fur- 
nish us the best information that we possess in regard to the 
strength of wooden columns. 

The summary of results is appended: — 



COMPRESSION OF WHITE-PINE POSTS. 



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COMPRESSION OF WHITE-PINE POSTS. 



66 I 













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662 



APPLIED MECHANICS. 



COMPRESSION OF WHITE PINE. — Single Sticks and Built Posts. 
In the multiple ones, dimensions of each stick are given. 







1 1 


Dimensions of Post. 




1 in 
Load 
. per 


Ultimate Strength. 


No. 

of 

Test. 


Weight. 


"■ss 

SIS' 

■so.s 








Sectional 
Area. 


Compressioi 
150 In. 
= 500 lbs 
Sq. In. 






u 

h-1 


.£3 

■3 




3 



<n . 

^ cr 

•JO) 




lbs. 




in. 


in. 


in. 


sq. in. 


in. 


lbs. 




664 


153 


11 


i77-5o 


4.48 


11.65 


52.2 


0.0545 


IIOOOO 


2107 


665 


143 


10 


180.00 


4.48 


11.64 


52.1 


O.IOIO 


81500 


1564 


666 


163 


5 


179.97 


4-47 


11.63 


52.0 


0.0895 


70000 


1346 


667 


228 


13 


180.00 


5-4° 


11.30 


61.0 


0.0505 


160000 


2623 


668 


193 


5 


179-93 


5.61 


n-73 


65.8 


0.0622 


156300 


2375 


669 


253 


5 


180.00 


5-64 


11.76 


66.3 


0.0608 


152300 


2297 


638 


IS!" 


ii 


180.00 
180.00 


4-5o 
4-5° 


11.60 
"•59 


5SI-" 


( 0.0670 ) 

1 0.0750 i 


200000 


1916 


639 
640 


I3il"» 


IS 


180.00 
180.00 
180.00 
180.00 


4-52 
4.49 
4-53 
4-52 


11.66 
11.62 
n-59 
"■59 


52} »M 

5.11-'- 


{ 0.0645 1 

( 0.0670 ) 

i 0.1060 1 

(0.0955 j 


212000 
149000 


2021 

1419 


642 


is I* 


\t 


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179.98 


5-57 
5-58 


11. 61 
11. 61 


&!■■»•♦ 


(0.0770) 
f 0.0390 J 


215000 


1661 


643 
644 


Isl- 




179.92 
179.92 
179.96 
179.96 


5.65 
5-6i 
5.60 
5.60 


11. 61 
11.62 
11..63 
11.62 


§:!!■*•» 
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( 0.0440 1 

( 0.0600 ) 

0.0596 

/ 0.0690 ) 


261000 
257800 


!995 
1980 


648 


!£!«•» 




180.00 
180.00 


5.60 
5.61 


11.72 
11.72 


i!:tt— = 


\ 0.0590 1 

{ 0.0700 ) 


268000 


2042 


649 

650 


i&l* 


i 9 
4 

11 


180.00 
180.00 
180.00 
180.00 


5-6o 
5-6i 
5-6i 
5-6i 


11. 71 
11.74 

"•75 
11. 71 


fell*** 


{ 0.0600 1 

t 0.0705 i 
i 0.0530 i 
) 0.0885 J 


277000 
240000 


2107 
1824 


645 


iaj* 


1? 


179.97 
179.97 


5-59 
5.60 


"•59 
11.60 


S3:5 1 *-»-• 


1 0.0560 ) 

1 0.0540 i 


263200 


2028 


646 
647 


i5l* 


12 

IS 


180.00 
180.00 
180.00 
180.00 


5-59 
5.60 
5.62 
5.62 


11. 61 
11.62 
11.62 
11.62 


6?:?i-3o-° 


i 0.0493 1 

} 0.0620 ) 
( 0.0630 1 
( 0.0700 ) 


249000 
248000 


1915 
1899 


678 


iS!«" 


11 


179.94 
179.94 


5-58 
5-57 


11.47 
"•45 


§:S i "" 8 


i 0.0529 1 

( 0.0642 ) 


245500 


1921 


679 
680 


13! » 


1- 

I 8 s 


180.00 
180.00 
180.00 
180.00 


5.62 
5.62 
5.60 
5-6i 


11.76 
11.72 
11.72 
"•73 


§11 '3-4 


\ 0.0664 I 
( 0.0705 1 
1 0.0650 1 

I 0.0495 $ 


249000 
278000 


1886 
2116 


663 


}3|«* 


ix4 

i 6 


180.00 
180.00 


5.60 

5.63 


"•75 

"•75 


sui^- 


( 0.0621 1 
? 0.0657 ) 


300000 


2273 


676 


IS!* 


I" 


179.94 
179.94 


5-6o 
5-6i 


11. 71 
"•73 


g1!'3-4 


< 0.0530 1 
' 0.0593 s 


274500 


2089 


677 


IS!* 


It 


180.00 
180.00 


5-6i 
5-68 


11.72 
11.72 


Sill-- 


\ 0.0551 1 

} 0.0625 ) 


255000 


»945 



COMPRESSION OF WHITE PINE. 



663 



COMPRESSION OF WHITE PINE.— Concluded. 
Single Sticks and Built Posts. 







1 1 


Dimensions of Post. 




Compression in 
150 In. Load 
= 500 lbs. per 
Sq. In. 


Ultimate Strength. 


No. 

of 

Test. 


Weight. 


£o.s 








Sectional 
Area. 






M 

c 


■6 




73 
a 

< 


In 

« . 

t-4c/2 




lbs. 




in. 


in. 


in. 


sq. in. 


in. 


lbs. 




690 


(175) 
< 226 > 600 

(199) 


(18 


180.00 
180.00 
180.00 


4-52 
5.56 
4.46 


11.62 
11.70 
11.62 


52-5) 

65.0 \ 169.3 
51.8) 


( 0.0460) 
< 0.0580 > 
( 0.0480) 


310000 


1831 


691 


jx6 4 ) 

( i59 ) 


Is 


179.98 
179.98 
179.98 


4.48 
5-56 
4-45 


11.60 
11.60 
11. 61 


52.0) 

64.5 [ 168.2 
5I-7) 


( 0.0526 ) 
I 0.0430 [ 
( 0.0390 ) 


372500 


2215 


692 


IS! 614 


ft 


i77- 2 5 
I77-25 
I77-25 


5.62 
4-SO 

5.60 


11.60 
11.60 
n-57 


65.2) 
52.2 > 182.2 

64.8) 


( 0.0580 ) 
< 0.0641 | 
( 0.0768 ) 


363000 


1992 


687 


218 536 


1" 

I 9 


j 80.00 
180.00 
180.00 


4-5o 
5-58 
4-52 


11.60 
11.62 

n-59 


52.2 ) 

64.8 > 169.4 
52.4) 


I 0.0460 ) 

0.0587 j 

(0.0533) 


325500 


1919 


688 


131 57s 


P 

(11 


180.00 
180.00 
180.00 


4-5° 
4.62 

4-5o 


11.60 
11.60 
11.60 


52-2) 

65.2 > 169.6 

52.2) 


(0-0565 ) 
< 0.0645 > 
( 0.0703 ) 


306000 


1804 


689 


[1881 

\ 203 \ 550 

( 159) 


H 


180.05 
180.05 
180.05 


4.48 
5.60 
4.46 


11.60 
11.62 

n-57 


52.0) 

65.1 > 168.7 

51-6) 


( 0.0510 ) 
I 0.0660 > 
( 0.0789 ) 


340000 


2015 


681 


M~ 


{ 9 


179-95 
179-95 
179-95 
179-95 


4-47 
4-52 
4.48 
4-5i 


11.60 
11.65 
11.65 
11.65 


5i.9l 
5 2 -5j 


ro.0700") 

! 0.0714 1 
] 0.0531 r 

to. 0762 J 


362000 


1734 


682 


!§}«■ 


(1 


180.00 
180.00 
180.00 
180.00 


4-5° 
4-50 
4-49 

4-5o 


n.63 
11.64 
11.60 
n.63 


52. 3] 

5 2 -4Uo 9 .i 
52.1 

52-3J 


(0. 0612") 
1 0.0546 1 
1 0.0542 f 

[0.0916 j 


414000 


1980 


683 


f l69 l 

ill 787 


11 


180.03. 
180.03 
180.03 
180.03 


4.46 
5.62 
5.60 

4 46 


n.63 
11.69 
11.70 
11. 61 


5i-9] 
5i-8J 


[0.0570! 
J 0.0530 1 
] 0.0494 r 
L0.0590J 


501000 


2133 


684 


§}" 6 


{» 

Lio 


180.00 
180.00 
180.00 
180.00 


4-5° 
5-64 
5.60 

4.48 


11.64 
n-59 
n.58 
11.61 


52-4] 
52. oj 


f 0.0664 l 
j 0.0610 1 
I 0.0548 f 
L0.0514J 


529000 


2255 


685 


r 145^1 


fio 

J 10 

h 


180.00 
180.00 
180.00 
180.00 


4-5o 
5-63 
5-6i 
4.48 


11.60 
11.62 
11.62 
11.62 


52.2") 

52. oj 


Co. 0645 ) 
1 0.0650 1 
] 0.0506 f 
L0.0546J 


430000 


1831 


686 


in*" 


(S 

1 12 
I 8 


180.00 
180.00 
180.00 
180.00 


4.48 
4-5° 
4-5o 
4-52 


11. 61 
n.56 
n.36 
11. 61 


52-0] 

52-5J 


f 0.0500I 

1 0.0315 ! 
I 0.0543 r 
^ 0.0680J 


395000 


1903 



664 



APPLIED MECHANICS. 













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CN 


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00 


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8 


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O 
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0006 

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0006 


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co 


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rj- 


lo r^. 


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r^. 


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NO 


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M 


d 


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00 n 6 


t> 


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t 


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co 




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lo no 


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p _ l 


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M 


% 


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NO 


NO 


NO 


NO 


NO 


LO 


NO 


LO 


LO 


LO 


LO 


LO 


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LO 


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LO 


LO LO 


LO 


LO 


LO 


LO 


LO 


LO 


LO 


LO 


LO 


LO 


LO 


LO LO LO 


LO 


»o 



COMPRESSION OF YELLOW-PINE POSTS. 



665 



g 

fa 

I 


Defl. horizontally. 
« «« 

" upward. 
" diagonally. 
" horizontally. 

« « 

" diagonally. 
" horizontally. 
Failed at knot 30 in. from end. 

" knots 8 " " " 

* " 36 in. from end, grain wavy. 
" at knot 17 in. from end. 
" ■ " " 15 and 29 in. from end. 
" " " 45 in. from end. 

" " near middle. 
" " " 50 in. from end. 


W> 
C 
V 

CO 
0> 

J 


is 


vDOONOr^Nror-)OOOvOr-»Tj-ONNMi-iO 

Tj-^-r^r^rovovOvOoocxD moo -t rovo N n \fl nvO 
On noo OS ts. Ovoo -1 fO m ■* n O O nvO >-> r^ Ov vO 


3 



< 


00000000000000000000 
.00000000000000000000 

</>000«">00"">000"^0 L O00 L O0"'>00*o 

£ r^ ro vo n n Q\ ^ O On •* O vo to ^t C\ O r^Tj-vo ■<*■ 

00 N "1\0 lOtouovO COrfVOCOOO O OS ""> O 00 C\ N 

m ci n n n n to « n n 


"(3 

CO 


c * oo »ooo n q\ q oq « n m oo oo q\ rp q\ vq vo <o\o o 
— oS« d\ w Cs 6 o\ oo od od ih^n^~>^n^n^h->i-ihih\n 
cj. n con con con n n n r^.r^r^r^r-*r^r-^r^r^t^ 

VI 


1 



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co 


■s 

0. 
a 

Q 


vO Os Os h oo CT\ vO ON O O ") N m ro "■> »-o N -^- m tj- 

.S in io i^i vo io io io iflio»no\d^Osd>do\cKdd\Os 




m <t n \ovOvO vnwM m oo VO ^-"tf-oo "^00 "i m o 
c ' "<f vq "*vq T f T t T t f 9 f P r ^^tv.r>r>'>r>vq r^r^r^ 

— lOlOifllfliOlOiOiOiOiONNtsNIsNNNNN 


,g 

ti 

c 


0000 n O vO O ro-^-coo O M O ro-<^Tj-oo m O r>. 
c -NNrofOroro<OTt , «!r-'^->-<0>-jHHi-<)-<i-<i-iNp-p 

— d vd vd vd d d dvdvdvdododoo d d d ■^■■^•■'fod 

,*i o n n n ^vouinnn^vOvO O O O mcocovo 


Average 

Rate 

of 

Growth. 


we- | | | | | | | | | |\0*«nOsoooovOfO't 
u cu 


Weight. 


q q o q q q q q q q q q *? *? q q q q q q 
«5von oSvd d vndodvdvd •<$- >-' « rod dvdodvdod 

£oO m oo VO 1- fOi-i voro^ON CO vO ro-^-rfvO "1 n m 

p*Nh«NNNNNNN«-'wi-iNNNNNCOCO 


# 




Ovw^vO r^ 00 0\ O ONO <- N ro^j-OO G\ O ^ >n O t^ 

* ^f * ■* O O "- OS O O NNNNN00CO00M00 
i/-)U">ir»U")»J"l*OVO-'3- iO«-OVO«-OVO»010lOVO«Oioio 



666 



APPLIED MECHANICS. 











































OT 










































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c 










































M 


































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*w 


































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u 

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ca 














ri 






a 














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rt 




T3 oS 








c 
o 


3 


5 


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o 


2 


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^ 


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5 


C 


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cc 


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<u 






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<u 










Q 






fe 


3 




















fe 


Q 




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co 


ON VO 


j. 


LO NO 


CO 


CO 


O 


N 


vO 


CO lo 


M 


2g 


t> 


8 


vO 


Tf M 


m 




r^ 




CN 


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CO 


CN 


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NO 


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c 


$£ 




O 


5- 


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r^ 


CN 


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CN 


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ON 


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cn co 




^f 


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s 

w 
S 




U1 


rt- 


CO 


CO 


M 


CI 


C) 


CO 


CO 


Cl 


Cl 


CO CO 


CO 


M 


CO 


>* 


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CO CO 


_j 




g 


o 


8 


o 
o 


o 

8 


8 


o 
o 


8 


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8 8 


3 


u? 






o 


o 


O 


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o 


o 


O 


LO 


ON C\ 


LO 


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8 


o 


o 




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Cl 


-r 


r-N 


LO 


o 


LO 


O 


CO 


o 


m 


~tr 


Cl 


o 


LO 


Cl 


CO lo 




< 




X) 


o 


CO 


r^ 


CI 


C) 


ci 


'CN 




o 


00 


rf co 


r^ 


r^ 


-r 


LO 


CO 


N M 


5 




co 


co 


N 


N 


o 


N 


Cl 


Cl 


C) 


Cl 




CO CO 


Cl 


c^ 


co 


«<t 


■* 


Tf Tf 


C 




c 


co oo 


o 


CO 


t">. 


M 


CO 


_ 


O 


-* 


rN 


"1" •> 


vq 


-* 


LO 


Cl 


Cl 


O NO 


_c 


: s 




in 


3 


th 


'*■ 


CO 


LO 


4 


LO 


d 


d 


vd 


tA. r^« 


r^- 


CN 


C! 


CO 


CO 


rj- CO 


t 

( 


!<3 


t/1 


r^ 


r^ 


r^ 


tv 


r^ 


r^ 


r^ 


c^ 


t> 


o 


CO 00 


CO 


o 


O 


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o 


CO CO 


a 


) 










































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CO 


ON 


O 


CO 


CO' 


r^ 


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r>« 


LO 


'* 


vq lo 


o 


LO 


IH 


-t 


w 


lo O 






f>i 


*■ 


CO' 


LO NO 


r^ 


t^ 


x^ 


■<* 


Th 


CN 


LO 


co 


LO 


VO 


vq 


Cvl N 




C 


ON 


CN 


CN 


ON 


ON 


ON 


CN 


CN 


CN 


CN 


CO 


LO LO 


LO 


LO 


LO 


LO 


LO 


vd vd 


j 


Q 








































.g 




-f 


LO 


H 


vO 


M 


ON 


Cv 


o 


o 


LO 


vO 


O co 

NO NO 
LO LO 


LO 


o 


!_! 


vS 

vd 


M 


LO LO 


o 
ft 


"3 


< C 










vq 


NO 


so 






<* 


"t 
tv 


vO 

LO 


Cn 
VO 


vO 

vd 


vO 

vd 


co co 






"<* 


VO 


in 


00 


LO 


O 


O 


C) 


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CN 


o 


<8 o 


8 


LO 


8 


8 


Cl 


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u5 


x 


n 


N 


N 


N 


M 


Cl 


Cl 


Cl 


N 


CO 


Cl 


CO 


N 


<7 


HH O - 




M 

c 

5 




30 


CO 





c 


6 


-* 


^r 


4 


CO 


00 


CO 


d d 


d 


d 


d 


d 


^ 


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■ 


vO 


o 


o 


o 


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CO 


CO 


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LO LO 


LO 


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w 


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N 


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M 


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<fl s 






































II 


j 1 * > 


uo-- 


N 




1 


i 


1 


1 


1 




1 


1 


1 


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LO 


1 


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CO 


CO 


LO M 


!°i 


n 


M 


N 




















11 N 


** 




1-1 






1-1 


< 


o 








































- 




q 


LO 


q 


O 


q 


q 


q 


o 


q 


q 


O 


q q 


q 


q 


o 


q 


q 


O O 


' 


Si 


EA 


IH 


cK 


H 


d 


» 


IH 


M 


s 


|H 


t> 


LO 


LO IV. 


Cl 


<vi 


-t- 


r^. 


(H 


M LO 






-Q 


CO 


ro 


vO 




CN 


O 


o 


CO 


Cl 


CO 


r-. on 


CO 


co vO 


LO 


ON NO 00 


i 


ss 




CO 


CO 


Tf 


<* 


CO 


LO 


LO 


LO 


lo vo 


LO 


CO CO 


co 


Tf 


-t 


LO 


Tl- 


LO LO 


6 V 






CO 


ON 


o 


M 


N 


N 


CO 


Tf 


r~^ 


CO 


ON 


t-i N 


CO 


Tj- 


=8 


z 


o 


M l-l 


fc 


o « 




CO 


CO 


CO 


CO 


CO 


Tf 


*t 


-* 


CJ 


M 


N 






co 




Tf r^ 


H 




LO 


lo 


LO 


LO 


LO 


LO 


LO 


LO 


LO 


LO 


LO 


t-* (^ 


t^ 


tT 


r^ 


t> 


r^ vO no 



COMPRESSION OF YELLOW-PINE POSTS. 



667 



Manner of Failure. 


<u 

CO 

<u 

+-> 

«+-! 
O 

"cO 

<u 

C to 



CO 


Defl. horizontally. 
( " " knots • 
( on concave side. 
(Defl. horizontally, ini- 
( tial bend in post. 
(Opened shakes and sea- 
( soned cracks. 


u 

in 

V 

5 


3^ 


- to r^. 

VO N O 1-1 
C\ O Ov vo 
ro i-H M ci 


"c3 
3 


< 


lbs. 
5260OO 

175OOO 
3IOOOO 
380OOO 


Sectional 
Area. 


B - (» N G\ n 
•*" N rv. vd to 
jf. ro O O tt 


(2 

V 

in 


U 

Q 


O to N ro 
n On r-^ 00 

.5 vd *o 1^1 vd 




O "^ CO 

c - n r-. 00 O 

— 00 vd vd 00 


3 


ro r^. to Tf 

-• CN N N N 

- « d 


Average 

Rate 

of 

Growth. 


c/i C 

tf- CO 1 1 1 

■eg, M 


i 
1 


lbs. 
578.O 

434-o 
450.0 
705.0 




0^ 


n to r^ vo 
r^. 00 00 00 
vO "* tJ- <*■ 



s «* 



66S 



APPLIED MECHANICS. 



COMPRESSION OF YELLOW PINE. 
Single Sticks and Built Posts. 



No. 

of 

Test. 



Weight, 
in lbs. 



sssa, 



Dimensions of Post. 



Sectional 
Area. 



CD ** 



g<0 W • 



Ultimate Strength. 






673 

A 73 * 
674 

675 
490 
67O 
489 

654 
655 

6g6 

651 
652 

653 
657 
658 
659 
660 
661 
662 

693 

694 

695 



in. 
180.05 
20.00 
20.00 

180.00 

180.00 

180.00 

180.00 

180.08 

180.00 
180.00 

179-93 
179-93 
180.00 
180.00 

180.00 
180.00 

180.00 
180.00 

180.00 
180.00 

179.96 
179.96 

179.98 
179.98 

180.00 
180.00 

180.03 
180.03 

180.00 
180.00 

180.00 
180.00 

179.97 
179.97 
179.97 

180.00 
180.00 
180.00 

180.00 
180.00 
180.00 



4.09 
4.01 
3-95 
4-Si 
4-34 
5 -05 
5.65 
5.85 

5.63 
5.62 

5.64 
5-63 
5.61 
5-6i 

5.58 
5-58 

5.58 
5.58 

5.63 
5-59 

5.63 
5-64 

5-59 
5-59 
5.61 
5-6i 
5.61 
5-63 
5.66 
5.60 

5.61 
5.61 

4-5° 
5-5o 
4.49 



m. 

"•35 
4.01 

3-97 

11.60 

11.60 

12.10 

11.74 

12.05 

11. 71 

11. 71 

11.72 
ii.72 

11. 71 
11. 71 

11. 71 

11. 71 

11.70 
11. 71 

11. 71 
11.68 

11.72 
11. 71 

11. 71 
11.72 

"•73 
"•73 
11.22 
11.24 

11.70 
11.72 

"•75 
"•75 
11. 61 
11.56 
11.62 

"•35 
11.36 

"•35 

"•35 
"•34 
"•35 



sq. in. 
46.04 
16.08 
15-68 

52-30 
5o-3o 
61.10 
66.30 
70.50 
65-9 



65.9 
66.1 
66.1. 

65-7 
65-7 

65-3 
65.3 

65-3 
65.3 
65.9 
65.3 
66.0 j 

66.0 1 

65-5' 
65-5 
65-8. 
65.8 

62.9 , 
63.3 ! 
66.2 
65.6 

65.9 
65.9 
52.2 
63.6 
52.2 



5i-o) 
59-o> 
5i-i) 



131. 8 
132.2 
i3i-4 
130.6 
130.6 
131-2 
132.0 
131-0 
131-6 
126.2 
131. 8 
131-8 
168.0 

165.2 

161. 1 



0.0370 

0.0492 
0.0490 

0.0455 

0.0418 
0.0540 

0.0292 
0.0315 

0.0368 
0.0466 

0.0620 
0.0514 

0.0395 
0.0360 

0.0559 

0.0550 ; 

0-0375 1 
0-0305 : 

0.0320 I 

0.0436 ! 

0.0312 1 
0.0372 ] 

0.0325 1 
0.0400 ' 

0.0410 1 
0.0365 ! 

0.0540 1 
0.0500 1 

0.0320 
0.0325 
0.0148 

0.0500 ' 
0.0650 ] 
0.0610 , 

0.0429 
0.0290 
0.0410 



142200 

94000 
99600 

131500 

121200 

230000 

205900 

250000 

470000 
580000 
480000 
360000 
588500 
436600 
580000 
448000 
600000 
510000 
410000 
388000 

564000 



474000 



3065 
5846 
6352 
2515 
2410 
3764 
3106 
3546 
3566 

4387 

3653 

2756 

4506 

3328 

4394 
3420 

4559 
4041 
3111 
2944 

33S7 

3027 

2942 



COMPRESSION OF YELLOW PINE. 



669 



COMPRESSION OF YELLOW PINE.— Concluded. 
Single Sticks and Built Posts. 







"o 


Dimens 


ions of Post. 




Compression in 
150 In. Load 
— 500 lbs. per 
Sq. Inch. 


Ultimate Strength. 


No. 

of 

Test. 


Weight, 
in lbs. 


22& 

SO.S 








Sectional 
Area. 






bo 

G 

h3 


M 
•S 

fe 


& 


3 
1 


5 c 

CW-i 

h-3 








in. 


in. 


in. 


sq. in. 








696 


J 253 j 660 
(190 J 


(is 


180.00 
180.00 
180.00 


4-5i 
5-50 
4.48 


11.24 
11.23 
11.23 


61.8 j 162.8 


( o.o337 ) 
j 0.0567 \ 
( 0.0715) 


480000 


2948 


697 


(242) 
j 249 J 706 


I 3 

(is 


180.00 
180.00 
180.00 


4-52 
5-43 
4-47 


11.60 
11.60 
11.58 


52.4) 

63.0 > 167.2 
51-8) 


( 0.0240 1 
J 0.0330 [ 
(0.0336) 


540000 


3230 


698 


J224) 
j 255 \ 775 
(296) 


I 1 

(15 


179.94 
179.94 
179.94 


4.46 
5-53 
4-50 


11.60 
11.70 

11-59 


517) 

64.7 J 168.6 
52.2 J 


( 0.0290) 
] 0.0385 [ 
( 0.0341 ) 


544000 


3227 


488 


911 


- 


180.20 


16.88 
{6.7a 


15-75 
15.75 


I08 -3 6 I 214 2 
IO5.84 \ 2I4 ' 2 


- 


700000 


3268 



On page 668 will be found a series of tests of spruce columns 
made in the Laboratory of Applied Mechanics of the Massachu- 
setts Institute of Technology, these columns having their ends 
resting against the platforms of the testing-machine. On the 
same page will be found also a series of tests made at the same 
place on timber columns with one end resting against a timber 
bolster. 

A perusal of this table will show a great decrease in strength 
due to the presence of the bolster. 

The four diagrams on the left of page 669 represent all 
the tests, with central load and no bolsters, of the four woods 
named, the abscissae being ratios of length to least diameter 
and the ordinates breaking-strengths per square inch. 

By whatever curve we may attempt to represent the values 
to be used in practice, it should pass nearly through the lowest 
results, as the timber was all of at least fair quality. It is also 
evident that up to a certain ratio of length to diameter, not 
far from 25, the strength is not affected by the length, and 



670 



APPLIED MECHANICS. 



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X Q 



COMPRESSION OF YELLOW PINE. 



67I 



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_, ff|- 


— FH+FF 



672 



APPLIED MECHANICS. 



hence this part of the curve should be a horizontal line, its 
ordinate being about 3500 for yellow pine, 2000 for white pine 
or spruce, and 3000 for white oak. 

For larger ratios than 25 the ordinate decreases, and might 
well be represented by some curve similar in character to those 
on page 425. 

The right-hand diagrams are due to Mr. E. F. Ely, and 
represent the lowest results of the Government tests on yellow 
pine and white pine, and also the values he proposes for use in 
practice. Mr. Ely's diagrams are represented by the following 
rules : Let A = area of section in square inches, f = constant 

given in the tables following, — = ratio of length to least side 



of rectangle. 

Then breaking strength = fA. 



White Pine. 


Yellow 


Pine. 


/ 


/ 


/ 


/ 


r 




r 




O to 10 


2500 


O to 15 


4000 


10 to 35 


2000 


15 to 30 


3500 


35 to 45 


1500 1 


30 to 40 


3OOO 


45 to 60 


1000 


40 10 45 


2500 




1 


45 to 50 


2000 




1 


50 to 60 


1500 



In the case of spruce and white oak, if it is desired to ap- 
ply the results to greater ratios of length to diameter than those 
tested, a similar reduction can be made in the value of f\.o that 
which takes place here in the case of white and yellow pine. 

§ 238. Factor of Safety. — Whereas, in the case of iron 
bridge-work, it is very common to use a factor of safety 4, the 
apparent factors of safety that have been used and recom- 
mended for timber have varied very greatly, and naturally so, 
because the values assumed for breaking-strength have been so 



TRANSVERSE STRENGTH OF TIMBER. 673 

very variable, and while some have advised the use of apparent 
factors of safety greater than 4, nevertheless most of the build- 
ing laws only require an apparent factor of safety 3, while 
making use of values of breaking-strength deduced from tests 
of small pieces. 

In view of the above facts, it is true that the values of work- 
ing-strength used in many cases have been very near the actual 
breaking-strength ; and, indeed, it would be impossible to 
recommend any suitable factor of safety to be used with re- 
sults derived from tests of small pieces. But if the true values 
k of the breaking-strength as derived from tests of full-size pieces 
be used, it would seem to the writer that a factor of safety 4 
will be sufficient for most ordinary timber constructions ; i.e., 
that we should use for working-strength per square inch one- 
fourth of the breaking-strength per square inch. In the case 
of mill-work, and in other cases where there is the jarring of 
moving machinery, it is advisable to use a somewhat larger 
factor. This same reasoning will also apply to the case of 
beams bearing a transverse load, where they are designed with 
reference to their breaking-weight. 

§ 239. Transverse Strength of Timber.— The table 
of Rankine, already given, represents the values of modulus 
of rupture as obtained from small specimens. Other 
values, not differing essentially from these, are given by Hat- 
field, Laslett, Thurston, Trautwine, and others, all based upon 
tests of small pieces. Confining ourselves to tests of full-size 
pieces, we find an account of a set of tests attributed by D. K. 
Clark, in his " Rules and Tables/' to Edwin Clark and C. Gra- 
ham Smith, The results are given below, and it will be seen 
that they are very much below those given by experimenters 
on small pieces. 



674 



APPLIED MECHANICS. 



Kind of Timber. 


Breadth 
and 

Depth. 


Span. 


How 
Loaded. 


Breaking- 
Weighti 


Modulus 

of 
Rupture. 




in. 


ft. 








American red pine 


I2.0 X 12.0 


15.OO 


Centre 


33497 


5238 


« <( << 


12.0 X 12.0 


15.OO 


« 


29908 


4680 


" " " 


6.0 X 6.0 


7.50 


" 


7370 


4608 


Memel fir . . . 


13-5 x T 3-5 


IO.50 


Distributed 


68560 


5274 


it u 






l 3-5 x 13-5 


IO.50 


" 


68560 


5 2 74 


Baltic fir . 






6.o X 12.0 


12.25 


Centre 


I9HS 


4878 


<< <« 






6.0 X 12.0 


12.25 


" 


23625 


6020 


Pitch pine 






6.0 X 12.0 


I2.25 


« 


23030 


5868 


« « 






6.0 X 12.0 


I2.25 


a 


23700 


6048 


<< << 






14.0 X 15.0 


IO.50 


a 


134400 


8064 


M « 






14.O X 15.0 


IO.50 


" 


132610 


7956 


Red pine . 






6.0 x 12.0 


12.25 


<< 


16800 


4284 


u U _ 






6.0 X 12.0 


12.25 


a 


19040 


4860 


Quebec yellow pine 


14.0 X 15.0 


IO.50 


Distributed 


68600 


4122 


(< <« <« 


14.O X 15.0 


IO.50 


« 


68600 


4T22 


" " " 


14.0 X 15.0 


IO.50 


Centre 


85792 


5148 




14.0 X 15.0 


IO.50 


" 


76160 


' 4572 

.. J 



Two tests by R. Baker are also mentioned by D. K. Clark. 

Bauschinger also made quite an extensive series of tests of 
German woods, an account of which will be given later on. 

A great many tests of the strength and stiffness of full- 
size beams of spruce, yellow pine, oak, and white pine, both 
under centre loads and distributed loads, have been carried 
on in the Laboratory of Applied Mechanics of the Massa- 
chusetts Institute of Technology. Tests have also been made 
upon the effect of time on the stiffness of such beams, also 
on the strength of built-up beams, and of floors and fram- 
ing-joints, all full size. A summary of the results obtained 
will be given, and conclusions drawn as to the proper values 
of the modulus of rupture and modulus of elasticity, etc., to 
be used in practice. 



TRANSVERSE STRENGTH OF TIMBER. 



675 



SUMMARY OF THE TESTS. 

The tests recorded may be divided into six classes: — 

i°. Spruce beams. 

2°. Yellow-pine beams. 

3°. Time tests. 

i°. Spruce Beams. — Before giving a summary of the tests 
made in this laboratory, I will insert some of the moduli of 
rupture and moduli of elasticity given by different authorities. 

Moduli of rupture are given as follows : — 



4°. Oak beams. 

5°. White-pine beams. 

6°. Framing-joints. 





Maximum. 


Minimum. 


Mean, 


Hatfield .... 


12996 


7506 


9900 


Rankine .... 


12300 


9900 


1 1 IOO 


Laslett 


9707 


7506 


9045 


Trautwine .... 


- 


- 


8lOO 


Rodman . . . . 


- 


- 


6168 



Hatfield's, Laslett's, Trautwine's, and Rodman's figures are 
from their own experiments. Trautwine advises, for practical 
use, to deduct one-third on account of knots and defects, hence 
to use 5400. The tables show the values obtained in these 
tests, and I will add a recommendation as to the values of 
modulus of rupture and modulus of elasticity suitable to use in 
practice. 

As a result of the tests thus far made in my laboratory, 
it seems to me safe to say, if our Boston lumber-yards are 
to be taken as a fair sample of the lumber-yards in the case 
of spruce, — if such lumber is ordered from a dealer of gooc 1 
repute, no selection being made except to discard that which is 
rotten or has holes in it, — that 3000 lbs. per square inch is all 
that could with any safety be used for a modulus of rupture, 
and even this might err in some cases in being too large ; 
(2°) that, if the lumber is carefully selected at any ooe lumber- 



6?6 APPLIED MECHANICS. 

yard, so as to take only the best of their stock, it would not be 
safe to use for modulus of rupture a number greater than 4000 ; 
and if we required a lot of spruce which should have a modulus 
of rupture of 5000, it would be necessary to select a very few 
pieces from each lumber-yard in the city. With a factor of 
safety four, we should have for greatest allowable outside fibre 
stress in the three cases respectively, 750, 1000, and 1250. 

The modulus of elasticity (i.e., that determined from the 
immediate deflections) was: maximum, 1588548; minimum, 
897961; mean, 1332500. 

As will be explained when the results of the time tests are 
given, if by means of the ordinary deflection formulae we wish 
to compute the deflection which a spruce beam will acquire 
under a given load after it has been applied for a long time, 
we should use for modulus of elasticity in the formulae not 
more than one-half of the values given above, or about 666300. 



TRANSVERSE STRENGTH OF TIMBER. 



677 



SPRUCE BEAMS. 







• c 
u 




•a 




afc 






4> 




re 


•O 


3 O. 






5 







O 


«« 


u 

h 






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c 
re 


ID 

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a 
a 

3 

Si 




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at .5 


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bo 

c 
15 
re 


S.SO 

3 V . 

•a ~ cr 
zm 


2; 


£ 


5 




S 


m 


2 




inches. 


ft. 


in. 




lbs. 




3 


2 X 12 


15 





Centre 


5,894 


5-526 


4 


2x9 


6 


7* 




7,322 


5,389 


5 


2x12 


15 







5,586 


5,237 


5<* 


2x12 


7 







8,982 




6 


2fX 9 


6 


8 


4 * 


7,586 


4.082 


7 


3X9 


4 





44 


11,086 


3,285 


8 


3 x 9 


10 





" 


6,086 


4,508 


9 


3X9 


15 





44 


8,086 


5.651 


10 


3i* 12 


20 





** 


6,586 


4.253 


11 


Hxi3l 


10 





(C 

4*ft. 

from end 


9,585 


3,787 


12 


13IX12 


16 


H 


\ 7,585 


3-271 


M 


7x2 


7 





Centre 


i,944 


8,748 


r 5 


i*x 6* 


7 







4,785 


7.562 


16 


3x9 


6 


8 


" 


9,9 S 5 


4,931 


17 


3x9 


6 


*\ 


4 points 
16" apart 


j- 16,744 


4,961 


18 


3.9X 1 


16 


°\ 


4* ft. 
from end 


[12.585 


5.218 


'9 


2X12 


14 





Centre 


4,404 


3.854 


20 


2 ,X 12 


14 





" 


5,108 


4.469 


21 


311X12 


14 





44 


8,627 


3,8j4 


22 


3^X 12 


14 





" 


12,545 


5.666 


23 


3l X I2i 


14 





" 


6,917 


2,995 


24 


3 xuf 


14 





" 


8,927 


5-442 


25 


2x9* 


14 





" 


3,i98 


4,139 


,6 


2^X 12 


14 





" 


6,819 


4-339 


27 


iilxio 


»4 





" 


4,3o6 


5,601 


28 


4^x12 


18 





44 


8,829 


4,8t6 


20 


4 Xl2i 


18 





14 


8,324 


4.586 


3i 


3|XT2 


18 





44 


7,721 


5,559 


35 


6 X12 


18 





" 


11,188 


4,196 


36 


2 XllJ 


7 


2 


" 


7,870 


3,599 


37 


4 xi if 


12 





" 


10,572 


4,135 


45 


3* xii* 


16 


4 


44 


8,072 


4,436 


46 


3TSXI2 


10 


2 


Centre 


13,772 


4,746 


49 


3* xiiJ 


14 





" 


12,076 


5,878 


60 


4x12 


17 


4 


12 points 


26,000 


7.448 


66 


4 X12 


15 


8 


Centre 


14,576 


7,211 


70 


4i X 12* 


17 


4 


12 points 


14,633 


3,748 


72 


4 XI2 T 3 B 


17 


4 




",333 


3,252 


90 


6 X12 


17 


4 


" 


26,100 


5, ,02 


196 


3*|xnH 


19 


10 


Centre 


9,187 


6,037 


i97 


3ix 11 
35 xii* 


18 


4 


" 


6,486 


4,757 


198 


17 


6 


" 


10,178 


5,993 


'99 


3*xiiH 


19 


6 


" 


9,784 


6,649 


200 


4 X 12f 


19 


10 


44 


6,597 


3,845 


201 


Slixii} 


18 


4 


44 


10.958 


5,94i 


202 


333 X 12 3^ 


18 


10 


44 


10,077 


5-946 


203 


,13„ , . 23 
3 1 6 x I I 35 


i9 


10 


It 


11,184 


7,626 


204 


3§? x 1 1 11 


18 


8 


44 


10,970 


7,066 


205 


3§5 x 123^ 


18 


8 


44 


8,082 


4,727 


297 


4 X 12 


15 


5 


44 


6,025 


2,923 j 


298 


3^X12 


9 


8 


44 


15,335 


4,700 


299 


3H x 12 


9 


8 


44 


M,i34 


4,484 1 



en tr 


3.5 3" 1 


- _Q 




Hi 3 


~%<s>\ 


c-> 




3 O* 


f <u j 




3.- u 


3 >,_§ 






o\s a 




2 





1,237,200 


.8.1 


1,067,900 


30I 




174 


938,500 


230 




308 




170 




I 4 I 






208 




126 




IO" 




304 1 






277 






465 






202 


1,482,600 


138 


1.588,500 


160 


1, 187,100 


137 


1,332,700 




898,000 


108 


1,572,500 




1,460,600 


123 


1,396,700 


155 


1,355,900 


167 


1,397,100 


134 


1,259,200 


129 


1,231,500 




1,347,900 






169 




133 




205 


1,461,000 


406 


1.336.200 


230 


1.551,800 


200 


1.228,600 


177 


1.587,600 




1,513,400 


153 


1,282.000 


120 


1.529,000 


170 


1,^10.900 


171 


1 . 100.200 


IOI 


1,490,200 


191 


1,475.100 


160 


1,850,600 


189 


1.470,400 


188 


1,208,000 


123 


1,010,000 


97 



Manner of Breaking. 



Crushing and tension. 
Tension. 



Tension and crushing 
Tension. 



Crushing. 
Tension. 



Tension and shearing 

Tension 

Shearing. 

Crushing and tension 

Tension. 



Crushing and shearing 

Shearing. 

Shearing and crushing 

Tension. 

Shearing. 

Tension and crushing. 

Tension. 



Shearing. 
Crushing. 
Tension. 



Crushing. 
Tension. 



,215,900 
,126,400 



232 



6 j 8 



APP 1. 1 F.D MECHANICS. 



SPRUCE BEAMS. 



en 


•o 


i 0. 
J3 3 


*£ bi 




Q. C/5 

3 £> • 


rtxi . 

5-1,2 


"B !/)* 

&& 




O 

o 


c . 

0*43 


111 


a 

4) "O 
C g 
c ° 

c3_l 


bo 

*S O 


o2£ 


•o'u V 

o\s a 


— c/3 O- 


Manner of Breaking. 


£ 


fc 







2 


OQ 


£ 


£ 


U 






inches. 


ft. 


in. 




lbs. 










3 00 


4 X 12 


9 


6 


Centre 


19,040 


5,652 


1,173,000 


298 


Tension. 


3d 


4 xi2$ 


9 


6 


44 


19,041 


5,537 


1,210,000 


295 


" 


304 


4 |XI2 


15 





tc 


8,475 


3,852 


1,466,800 




Shearing. 


305 4l5 XI2 


15 





" 


9,779 


4,515 


1,345,600 


i5 2 


Tension. 


307 4lB XI2 


16 





44 


10,233 


4,594 


1,238,100 


156 




308 


4ix 12 


16 





M 


12,186 


5,735 


1,466,700 


179 




3°9 


3TB x "fr 


16 





ts 


9,283 


4.618 


1,067,300 


141 


Tension. 


3" 


4 iXI2 


16 





it 


8,891 


4,184 


1,156,100 


129 


" 


3M 


3ixi2 


16 





*• 


6,670 


3,442 


978,700 


102 


tt 


316 


4 xnj 


16 





SI 


11,885 


6,068 


1,290,300 


184 


" 


3 J 7 


4rs*i2 


16 





El 


12,189 


5,638 


1,479,400 


i75 


44 


3i8 


4 ^ ii* 


16 





(t 


11,875 


6,113 


1,414,800 


191 


*' 


319 


3fxi2 


16 





M 


12,386 6.393 


1.470.900 


201 


Crushing. 


320 


3tI x "tI 


16 





H 


5,386 


2,828 


1,092,600 


85 




321 


3**«* 


16 





M 


13,086 


8,120 


1, 899,800 




Tension and shearing. 


322 


4 xi 2ib 


16 





it 


9,57i 


4,639 


1,081,400- 


i45 


Tension. 


323 


3 }XI2 


IS 





(C 


9,170 


4,585 


1,196,600 


i54 


44 


325 


4 XI2£ 


16 





«< 


8,i75 


4.003 


979,3 c o 


128 


44 


327 


4 X 12 


15 





« 


4,665 


2,187 


890,700 


74 




328 


3i XI1 * 


16 





l« 


12,375 


6,522 


1,701,500 


203 


Shearing. 


33° 


3 |XI2 


is 





44 


6,075 


3,037 


1,089,500 


103 




333 


4ix"il 


is 





44 


5.353 


2,512 


849,500 


83 




33 6 


4 XI2i 


15 





H 


7,96i 


3,655 


1,002,400 


124 


Tension. 


344 


3x|xi2| 


15 





tt 


10,761 


5,184 


1,369,100 


175 


" 


348 


3ixnf 


15 





tt 


13,162 


6,652 


1,689,600 


217 


" 


349 


3£xiii| 
sHxiTf 


15 





'* 


7,760 


3.801 


1.116,300 


126 


Tension at knot. 


35' 


15 





44 


10,464! 5.385 


1,382,000 


176 


Crushing. 


353 


3fxi2 


15 





44 


11,978 


5,989 


1,253,900 


200 


" 


354 


4 x I2 T T B 


16 





44 


8,184 


4,049 


1,169,000 


127 


Tension. 


355 


3t|xiiH 


15 





It 


7,760 


3,893 


1,233,400 


126 


44 


356 


3ixnJ 


15 





44 


9,963 


4,774 


1,464,700 


*57 


44 


359 


3 x 9i 


17 





44 


4,740 


4,958 


1,147,500 


120 


Compression and tension 


363 


4isxi2i 


15 





44 


10,170 


4,598 


1,285.100 


157 


tt tt «t 


365 
366 


3ixi2J B 
3 xgf 


17 
17 







9,176 
5,540 


4,980 
5.793 


1,129,400 
1,411,200 


147 
141 







16 


Compression and tension 


367 


2 fx 9 i 


17 





44 


1,630 


1,869 


1,211,000 


47 


Tension. 


369 


3x|xi2i 


17 





44 


9,876 


5,217 


1,373,000 


157 


Tension and compression 


37o 
37 2 


4 X 12 
3**9* 


17 

12 







10,789 
[ 3,722 


5,732 
1,794 


1,408,700 
854.100 


170 

94 




A 


2 points 1' 
from centre 


Tension. 


373 


3 *9| 


12 


-\ 


2 points 2' 
from centre 


[ 9,595 


4,733 


1,126,500 


244 


Tensionand compression 


374 


zfxgf 


12 




2 points 2' 
from centre 
i at centre 


( 10,767 
j 


5,673 


1,593.000 


289 


Shearing. J 


377 


2*3*9* 


12 


H 


J— 3' each 
from centre 


V 4,626 


2,722 


1,072,200 


128 


Tension. 


378 


3 £XI2j 


15 


H 


2 points 2' 
from centre 


p7,259 


6,252 


1,720,000 


277 


Tension. 


380 


itSxqH 


12 





Centre 


4,216 


5,086 


1,361,000 


176 


Tension. 


383 


3fXI2g 


16 





44 


9,474 


4,840 


1,205,000 


153 




384 


2£X IO 


II 


6 


44 


9,36o 


6,739 


1,537,000 


244 


Compression. 


385 


2 J X IO 


II 


6 


'• 


5,o°5 


3,604 


1,082,000 


I3 1 


Tension. 


387 


4ixi2i 


l6 





44 


9,869 


4,735 


1,666,600 


149 


Tensionand compression 


387 a 


4 xnf 


l6 





it 


ii,956 


6,232 


1,442,800 


191 


Shearing, crushing& ten. 


389 


2|XQ| 


II 


6 


44 


7,702 


5,945 


1,445,000 


213 


Tensionand compression 


39 2 


. 3^X12 


l6 





ti 


15,813 


9,036 


2,132,400 


284 


Tension. 


395 


3«x«ii 


12 





tt 


8,132 


3,339 


1,055,500 


138 | Shearing. 


397 


4 X 12 


17 





44 


9*574 


5,084 


1,102,000 


151 i Tension. 



TRANSVERSE STRENGTH OF TIMBER. 



679 



SPRUCE BEAMS. 







a 




fci 




- -. 


.2 « 


gS-d 






£ 






a 




3 a 

■"-* 3 


s a 






O. 


5 




■3 




O.CT 


rt 


5 c c 






p 


"5 

.0 




3 


1 


3tn 


«-n-3 c 


«- j 


1 


«5 




•a 

c 


u 

y 

c 
cd 


•- 


— 


u 

V 

a 


bo 
c 
13 


a 


£ 
« a <u 

3 3 


•a g § J Manner of Breaking. 

.5c/>C/5 
3 u 


"3 




3 


c 


as 


•3 e c 
O.S£ 


•a >,o* 


■- - « 




5 i 


01 


S. 


?j 


ti 


0. tic/5 


■3 ° a 




Q 




£ 


n 


£ 


§ 


U 






inches, ft. 


in. 




lbs. 










398' 3**12 17 





Centre 


10,677 


5,854 


1,602,000 


174 


Tension. 


400 2^f x io£ 


*5 





" 


5,725 


5,36i 


1,433.400 


152 




401 ifrxgj- 


15 





" 


4,107 


6,221 


1,609,600 


170 


4i 


404 3 XIO 


15 


6 


" 


6,828 


6,350 


1,563,900 


J 73 


" 


405 4 X 12 


is 





II 


7,160 


3,356 


1.244.300 


"3 


Shearing. 


407! 4^x1 if 


16 





" 


8,660 


4,287 


1,099,300 


134 


Tension. 


408 4 A I2 T 3 g l6 





" 


8.762 4.248 


1,280.300 


136 


" 


409 3ll X 12 15 


6 


" 


6,152! 3,127 


986.000 


102 




4 II 4 X 12 I 7 


6 


12 points 


9.157! 2,645 


1.117,000 


J 43 


Tension. 


414 2l|XTO 15 





Centre 


5,200 4.902 


1.270.500 


138 


Crushing at knot. 


415 3 x Qt§ j 6 





" 


3.700 


3,723 


1,105,700 


98 


Tension. 


i 416 2|x 9 i| 15 


6 




7,900 


8,087 


1,548,400 


215 


Longitudinal shear and ten- 
sion. 


417I- 4±xn£ 14 





" 


12,100 


5,139 


1,622,500 


182 


Tension. 


421 


3$*"* 15 





" 


12,000 


6,010 


1,318,000 


199 


Tension and longitudinal 




















sAear. 


422 


3i*i:f 


14 





" 


13,200 


6,182 


1,304,700 


217 


Tension. 


423 


4 X12 


14 





u 


13,260 


5,762 


1,213,100 


207 


Tension, crushing, and 
longitudinal shear. 


424 


3*xn£ 


16 





" 


12,000 


7,105 


2,100,000 


221 


Tension. 


425 


4 x 12 


14 





" 


9,710 


4,206 


1,067,800 


152 




426 


4£*I2 


14 





« 


6,160 


2,562 


966,880 


93 


" 


427 

428 

429 


4 X12J 


14 







8,110 


3,663 


1,249,900 


135 


" 


4 X 12 


14 





II 


8,665 


3.753 


1,148,300 


135 


" 


43° 


2|X IO 


14 





" 


4,440 


4.217 


1,377,600 


127 


" 


431 


zfxio 


14 





" 


6,500 


5,897 


1,464,200 


177 


11 


432 


2* XIO 


M 





" 


5,500 


4,99i 


1,156,400 


150 


" 


433 


4fxioi 


14 





iC 


2,925 


2.542 


896,800 


78 


" 


434 


2^X10 


14 





" 


7,020 


6,525 


1.819,900 


196 


" 


435 


2$X IO 


14 





" 


4,990 


4,5i8 


1,180,300 


I3 6 


" 


436 


2ix 9 || 


14 





(« 


5,120 


4,830 


1,242.800 


i39 


" 


438 


2}X 9 | 


14 





" 


6,400 


5-969 


1,358,100 


i77 


" 


439 


2jx IO 


14 





" 


6,000 


5,467 


1,263.700 


164 


" 


441 


2|X IO 


14 





" 


8,080 


7,347 


1.732,800 


220 


" 


442 


4|x9t b b 


it 





" 


4>!3o 


4-5 T 9 


1,368,000 


166 


" 


443 


4 X12 


14 





" 


10,760 


4.780 


1,129,800 


172 


" 


444 


4 X12 


14 





" 


10,000 


4,495 


1,238.200 


163 


" 


445 


3^X12 


14 





N 


7,875 


3,673 


1,327,600 


i33 


Longitudinal shear and ten- 
sion. 


446 


4x12 


14 





U 


10,400 


4,633 


1,286,800 


167 


Tension. 


448 


4^X12^ 


H 





K 


10,800 


4.639 


1 236,400 


167 


" 


449 


3*XI2 


14 





" 


8,500 


3-929 


1,213.000 


142 


" 


450 


4&XII* 


14 







10,400 


4,658 


1,160,900 


166 


Longitudinal shear and ten- 
sion. 


456 


7Hxi2t 


19 





" 


14,000 


4,098 


1,267,000 


"3 j 


Tension. 


457 


7*XI2 


19 





" 


14,750 


4,456 : 


1,176,700 


121 


" 


458 


7ixi2 


19 







16,950 


5,216 


1,440,000 1 


139 ! 





68o 



APPLIED MECHANICS 



SPRUCE BEAMS. 







a 


bib 




<u v 


■J tu 


£? in 




05 

u 

H 


Xi 

Q. 
V 

Q 
•u 

c 
a 

JS 


u 


c 

S 
a 


-1 


u 

V 

c 


•0 

rt 


bo 
c 
15 


3 rt 
fl-J-g 


•s a 

(0 

hi 

3 3 


3 u 


Manner of Breaking. 






•S 3 


a 




Sag 


•a >%cr 






o 


* 


.56 73 


rt 




0. tic/3 


73 a 




£ 


: 


3 


m 


s 


s 


u 






inches. 


ft. in. 














459 


6 X12 


14 Ce 


ntre 


n,6oo 


3.345 


915,000 


123 


Tension. 


460 


5i*i;£ 


14 




!6,35o 


7,58o 


1,640,800 


184 


Crushing, t.ension,and shear 


461 


6 x 12 


14 




15,600 


4,59o 


1,369,000 


166 


Shear. 


481 


5**12 


15 




21,000 


6,120 


1,845,000 


231 


" 


488 


6JXI2 


19 ' 




12,950 


4,900 


1,271,000 


130 


Tension. 


489 


6 xnf 


22 ' 




11,700 


5,800 


1,549,000 


129 


Crushing and tension. 


491 


5 f-XI2 


14 ' 




13,500 


4,070 


1,117,000 


147 


Tension. 


495 


5 JXI2 


14 




15,500 


4,660 


1,195,000 


168 


" 


496 


6 X 12 


10 ' 




22,500 


4,7oo 


1,186,000 


236 


Shear. 


497 


6 X 12 


10 ' 




24,600 


5,150 




258 


'* 


498 


6 X 12 


15 




5.55o 


1,850 


923,000 


60 


Tension. 


499 


6 X 12 


15 




16,700 


5,230 


1,274,000 


176 


" 


5°i 


6x12 


16 




10,800 


3,640 


948,000 


"5 




j 


\verage vs 


dues 






4,5*i 


1,310,584 





MAPLE BEAMS. 















• 










a 


bs 






.2 « 


rt u 






SI 

a 

V 

Q 

•v 

a 
rt 
X! 


4> 


c 




3 rt 


a a 


V (S 




tfl 

u 

H 


u 

V 

x> . 

(0 

"J 


•a 

•J 

"o 

u 

a 


•0 
g 

ho 
c 
!3 


a 

"* rA 


rt 

lag 

3 3 


to 

a 

13 -S 


Manner of 
Breaking. 




•o 


2 3 
.<2c/3 


c 

rt 


w 


o.£° 


■a >,cr 

O.^t/3 


« c c 




£ 


s 


P 


s 


PQ 


s 


s 








inches. 


ft. in. 














463 


4 x 12 


15 6 


Centre 


9,650 


4,732 


1,396,0*0 


155 


Tension. 


467 


4 xi2 


16 




12,300 


6,200 


1,627,000 


196 




469 


4 X12 


13 




17,800 


7,280 


1,448,000 


282 




470 


4 xii} 


14 




9,200 


4,260 


1,262,000 


150. 




473 


3^x12 


15 




14,600 


7.900 


1,587,000 


265 




475 


3**"* 


14 




18,450 


8,570 


1,597,000 


305 




477 


2|X 11$ 


14 




15,500 


11,080 


1,702,000 


385 




A^ 


perage vali 








7,146 


1,517,000 











TRANSVERSE STRENGTH OF TIMBER. 



68 I 



Yellow-Pine Beams. — The moduli of rupture in common 
use are given as follows by different authorities ; viz., — 





Maximum. 


Minimum. 


Mean. 


Hatfield .... 
Laslett 

Trautwine .... 

Rodman .... 


2II68 
14162 

9876 


9OOO 
IOO44 

C Yellow pine 
( Pitch pine 
8796 


I53OO 

12254 

9OOO 

9900 

9293 



A summary of the figures obtained from these tests will be 
given in a table at the end of these remarks. 
It will be observed that we have for 





Maximum. 


Minimum. 


Mean. 


Modulus of rupture . . 
Modulus of elasticity 


II360 
2386096 


3963 
II62467 


7486 
I7579OO 



If by means of the ordinary deflection formulae we wish to 
compute the deflection which a yellow-pine beam will acquire 
under a given load after it has been applied for a long time, 
vve should use for modulus of elasticity in the formulae about 
one-half of the values above, or about 878950 (see report of 
time tests). 

For the modulus of rupture of yellow pine of fair quality, 
in the light of the tables on pages 683 and 684, I should not 
feel justified in using a number greater than 5000 pounds per 
square inch. With a factor of safety four we should have 
about 1200 as our greatest allowable fibre-stress. 



682 



APPLIED MECHANICS 











YELLOW-PINE 


BEAMS. 






J3 


a 

V 




i 




§■8. 


6fc 


£* co" 

.- .a • 

C 1 ^ cj 








* 





1-1 


•a 



co c V 


a! 


Us 

1— 1 <u 




in 


•a 
a 

■a 


V 

u 

c 


«5 

O 

a 
a 

3 


O 

U 

it 


•J 
bo 
c 

3 


•g.J.5 

s ^ 3 


h 


Manner of Breaking, 




E 


.2 

Q 


/3 




OSOQ 
8 


O. ^C/3 








inches. 


ft. 


in. 




lbs. 










3° 


3 xi3J 


14 





Centre 


15,158 


6,614 


1,937,000 




Shearing. [shearing. 


32 


4tb x 12^ 


18 





" 


13,75! 


7,383 


1,734,000 




Tension, crushing, and 


33 


3i|xi2± 


18 





" 


9,832 


5,386 


1,794,000 




Shearing. 


47 


3 X'3* 


M 





" 


i9>574 


8,696 


2 386,000 


359 


Crushing. 


50 


4 XI4TB 


21 





" 


12,875 


5,9M 


1,256,300 




Shearing-. [shearing. 


53 


3ixi4 


24 


6 


" 


10,076 


7,206 


1,784,400 


179 


Tension, crushing, and 


54 


3 Xl2i 


24 





" 


9,576 


9,380 


2,Il6,8oo 


203 


Crushing. 


56 


3ixi 4 


15 


4 


" 


10,572 


4,764 


1,490,400 


185 


Tension. 


57 


2lfX!2 


*9 


2 


" 


8,472 


6,950 


I,444,.SOO 


182 


" 


59 


9 xi 3 } 


24 





11 


21,083 


5,352 


1,417,800 


133 


Tension and crushing. 


62 


4i X 12^ 


19 


10 




15,461 


9,102 


2,038,000 


237 


Crushing. 


63 


4tbXI2^ 6 


20 





" 


14*073 


8,i45 


1,599,000 


211 


Tension. 


64 


4ixi2i 


19 


10 


" 


10,573 


6,098 


1,918,000 


161 


" 


65 


4 Xl2i 


19 


8 


" 


",573 


6,782 


1,966,700 


183 


Crushing. 


67 


4ixi 2 


18 


6 


" 


13,374 


7,277 


1,787,600 


196 


Tension. 


68 


4 XI2^ 


19 


9 


" 


17,676 


10,872 


2,381,700 


278 


*' 


69 


3i% x 14 


20 





" 


6,675 


3,963 


1,169,300 


"5 


Crushing. 


7i 


4±XI2 


18 


2 


" 


16,074 


8,248 


1,512,200 


227 


Tension. 


74 


4 X 12 


20 





" 


11,071 


7,004 


I,628,IOO 


175 


" 


75 


4 xnf 


19 


9 


" 


i3,77i 


9,39i 


1,850,700 


233 


" 


76 


4ixi2 T 7 5 


17 


4 


12 points 


15,82s 


4,207 


I,344,IOO 


233 


" 


77 


4 ±XI2 


17 


4 


11 11 


37,325 


10,286 


2.123,200 


55i 


" 


78 
79 
81 


4 XI2^ 
4 XI2 

4iX!2} 


22 


10 


Centre 


7,172 


4,845 


T, 455,300 
2,087,600 
I,l62,500 


109 




*9 
17 


8 






4 


12 points 


16,025 


4,349 


"238' 


Tension. 


82 


4 XI2 


*9 


8 


Centre 


i5,57i 


9,671 


1,607,300 


246 


" 


84 
85 


4iXI2i 

4 xnf 


21 


4 
6 


" 


"•374 
16,874 


6,985 
11,360 


1,501.000 
2,246,200 


167 
271 




20 


" 


Tension. 


87 


4 X12I 


21 


4 


" 


11,272 


7,335 


1,535,600 


175 




88 


6 XI2* 


20 


4 


" 


J 5, 28 3 


6,112 


I,6l3,000 


161 


" 


9i 


4 X 12 


19 


10 


" 


18,074 


",303 


2,223,8oO 


282 


Tension and shearing. 


92 
144 


6x12 


6 


5 





38,090 
9,433 


5,092 
6,012 




397 
145 




4^X12 


21 


11 


I,628,IOO 


Tension. 


i45 


4ixi2^ 


19 


3 


" 


11,201 


6,400 


1,472,000 


191 


" 


146 


4 XI2 


19 


4 


it 


I3,34i 


8,060 


1,839,700 


206 


" 


147 


4 XI2 


19 


2 


" 


16,748 


10,032 


2,286,000 




Shearing. 


216 


4ixi 2 J 


*5 


8 


ii 


x 5,453 


7,040 


1,557,300 




Crushing and shearing. 


218 


4i x i 2 £ 


15 


6 


11 


15,613 


7,484 


2,010,300 


244 


Crushing and tension. 


219 


4ixi 2 | 


15 


6 


" 


16,632 


7,427 


1,623,800 


244 


Tension. 


220 


4&xi 2 £ 


15 


6 


" 


17.710 


7,982 


I,775,IOO 


265 


V " 


221 


41x12^ 


x6 





it 


16,330 


7,836 


1,931.200 


248 


" 


222 


4ixi2i 


16 





" 


18,515 


8,884 


1,786,000 


285 


" 


223 


4 x 12 


17 





ii 


13,492 


7,167 


1,638,500 


212 


Crushing. 


224 


Sllxiiil 


17 





" 


16,426 


8,958 


1,938,900 


264 


Tension. 


225 


4^X12 


15 


6 


k 


18,705 


8,786 


1,729,900 


285 


" 


226 


4£xi2 


15 


6 


" 


i6,594 


7,794 


I,6ll,200 


2 53 


" 


252 


,15 v , T 15 


17 





" 


11,006 


6,002 


I,27I,000 


177 


" 


254 


4 x 12 


16 





" 


15,226 


7,613 


I,6l7,700 


240 


" 


255 


4l 7 5Xl2t 


15 


6 


11 


19,425 


7,975 


2,214,400 


267 


Crushing. 


257 


4* x i2fc 


16 





ii 


17,424 


8.031 


1,789,800 


256 


" 


258 


4/s x 12 


16 





it 


18,319 


8,256 


1,830,000 


260 


Tension. 


261 


4&X12 


IS 





it 


17,818 


7,978 


1,989,700 


268 


14 


1 3 2 4 


3fXI2} 


15 





it 


12,510 


6,002 


I,7T9,500 


206 


Shearing. 


326 


4 X12J 


16 





«t 


14,011 


6,862 


I,98o,6bo 


219 


Tension. 


329 


4ix 13 i 


15 





" 


23,324 


9,971 


1,845,600 


340 


" 



TRANSVERSE STRENGTH OF TIMBER. 



68 3 



YELLOW-PINE BEAMS. 





• 


c 




bo 




3 S 

X 3 


6 fe 

Z a. 




(0 
V 

H 


u 
Q 
•a 
c 
cd 

•0 


u 

S! 

V 

u 

c 
aj 


U5 
l-c 

O 

a 
a 
3 


■5 
8 

u 

c 
a 


g 

bo 


Br " 
MS. 


<& . • 

JUS 

•0 >,cr 


•n }5 g ! Manner of Breaking. 

3 u 





In 


-0 


cd 


u 


o.S£ 


c «^/3 


-^ a 


2 


^ 


5 




£ 


m 


S 


s 


U 




inches. 


ft. 


in. 




lbs. 








33i 


3f *"£ 


15 





Centre 


13.709 


6.774 


1,619,800 


225 Tension. 


332 


4 *i3i 


*5 





" 


21,733 


1 8.356 


2,109,600 


310 Shearing. 


334 


j 3£xi 3 


15 







17.400 


i 7-i°5 


1,586,600 


259 Tension. 


335 


4x12 


T5 





" 


16.228 


7.686 


1,911.400 


257 Shearing. 


339 


| 4 xnif 


16 


6 


14 


15.7^7 


8.373 


2,128,800 


250 Tension. 


342 


4iV x l2 £- 


t6 





" 


9.616 


! 4.502 


1,380,500 


142 




346 


-i xi'l 


16 





" 


16,227 


8.284 


1.857,800 


256 


" 


347 


4 X12 


17 





*' 


13.536 


j 7,191 


2,041,000 


211 


" 


350 


4 > T2 


T 5 





" 


19,212 


1 8.912 


1.790.500 


299 o L 


352 


4i>< i2|- 


*7 







18,238 


j 9,087 


2,059 600 


271 Shearing. 


? : 7 


4ixi,f 







" 


22.546 


10.278 


2,595,600 


331 Tension & compression. 


358 


4ix.2| 


17 





" 


20.825 


10.371 


2.137.340 


3 12 „, 


360 


3 x 10 


T 5 





" 


12,0,1 


10,864 


2,122,000 


302 Tension. 


36. 


3 x 10 


15 





" 


7.264 


6,485 


1.632.700 


179 


362 


3 *9l 


r 5 





41 


8,467 


7,815 


1,833.700 


214 Shearing. 


364 


4 x 12J 


1 5 





" 


15,208 


6,879 


1,667,500 


237 Tension. 


368 


4 XI2| 


15 





" 


17,714 


8,133 


1,899,100 


276 j 


376 


3 fxii| 


15 





" 


14,712 


7,388 


1,800,300 


246 , " 


38i 


3*xit| 


15 





" 


15,201 


7,761 


1,566,400 


25S Tension and shearing. 


382 


4 X12J 


15 







17,525 


7,803 


1,882,700 


269 Compression. 


386 


4 X 12 


l6 







12.no 


6.055 


1,310,000 


191 Tension. 


388 


4 X12J 


16 





" 


'0.534 


5,059 


1.658,800 


162 Shearing and tension. 


39o 


4 X12^ 


'5 







22,100 


10.147 


2,234,800 


342 Shearing. 


39i 


4^x12 


17 





" 


18.391 


9>474 


2,068,200 


279 Tension. 


394 


41x12* 


16 





" 


19,713 


8.902 


1,096,400 


286 Shearing. 


396 


3^x12 


20 





" 


10,296 


7-354 


1,889.000 


186 


Tension. 


396a 


3*^12^ 


8 


6 


" 


25,827 


7,392 


2,217.000 


445 


Shearing. 


399 


4^X12^ 


16 





44 


8,602 


3.828 


1,306,000 


!24 


Tension. 


402 


4**«A 


15 





" 


15.192 


6.303 


1,542,000 


2I 5 


" 


403 


4**™* 


*5 







10.886 


4,636 


1,058,200 


158 


it 


410 


3fX!2f 


15 





2 points 2' 
from cen. 


18,590. 


6.304 


1,733,400 


297 


" 


j 412 


4JXI2J 


14 





do. do. 


15,865 


4,627 


1,774,100 


237 


Shearing. 


! 4i3 


3fxi2i 


J 5 





do. do. 


14.852 


5.069 


1,778,900 


236 


Tension. 


418 


4 X 12 


14 





Centre 


20, 500 


9.064 


2,278,500 


326 


" 


41 


4 XI2 


16 







12,850 


6,54i 


1,664,800 


207 


Crushing at top & shear- 
ing above neutral axis. 


420 


4 XI2 


14 


6 


" 


14,000 


6,442 


1,684,400 


225 


Tension. 


429 


4 X12 


14 







15,600 


6,759 


1,498,500 


244 


Tension, compression. & 
longitudinal shear. 


I 437 


6 x i6£ 


'7 


6 




38,000 


7.407 


1,777,700 


289 


Tension and shear. 


44: 


4X12 


16 





44 


19,500 


9.908 


1,778,400 


3 12 


Tension. 


447 


3ixi2i 


18 





44 


13,900 


7.908 


1,672,900 | 


227 1 


'• 


45 ' 


4 X12I 


18 


6 




9.3.SO 


5.446 


2.013.200 1 


T 5 J 


Longitudinal shear and 
tension. 


: 452 


4ixi 2l \ 


14 







i5,75o 


6,570 


1,704,900 i 


240 j 


Tension and longitudi- 
nal shear. 


| 453 


4 XI2 


M 





" 


13,000 


5.800 


1,804.000 


209 


Tension 


I 454 


4^x 12 


X f 





" 


12,000 


6.848 


1,603.800 


182 | 


Tension & compression. 


455 


4&XI2 


16 





" 


16, 750 


8.200 


1.656,000 


259 Tension. 


Average values 


7,442 


1,783,000 





684 



APPLIED MECHANICS. 







C TJ 




52 w P 






.~ w 




rt js — 






s! 




p s 






»-. 




)-. — ' OT 






1) 




« 0) 






.O. ** 




.O. rt ~ 




' 


P "* 
C OO 
3 O0 






10 




J " 




•J ^ i 


a 




1 «=> 




\ %£ ^ 




• ^ oo 




<N C M in O 




N V 00 




t* 




-oc 




H «J ^ O0 

= •£ s 1 £ 

*■• 5 v " « 
w p *> cu .2 

O v > V) j* 

£ O 


' 


■* 




O^OO 00 


O O O oo co 

co O o O r^ 


*t 


O rfCOOco ^-i^co Q O 

MTtf-MCOONMt^OO 

O^oo <n o>r^r»iN in o> o 


co c* a> 

OO N O 


r^ 


3 '3 


m O O rf co 


r^ 


sO O CO 


i^ co r^ tj- r^ 


«t 


r^ O co on co N inco 0\ 


£ s 

T? rt 


€ "t m 


O N N co in 


<t 


O O r» i^ n- r>^ co m r^. co 


N NOO 


co oo p< m o 


r^ 


OO O^O^COO M MOOOO 


O -5 








MM M MM 


s« 










10 u 

a s 










O o ■* 


O O O O rj- 


<± 


in t co rf O inoco co «i- 


Q\ vO OO 


in co CO rf m 


& 


Ot^O NOO CO CO COM IN 

co coo O^m inoeor^to 


»n O O 


O in in N N 


«* 


3 O. 


m o rf 


t^. in in co in 


in 


^oo inrfo coino rfm 


•O 3 










l« 










&„- 


O O M 


M M ^- M CO 


<+ 


inooo O O con Ot-^O 
00 O rtoo O incoooo O 


g J3 w 


t>. r» r^ 


r^ N CO N 00 


O 


.* .S°i2 


i-> o o 
co O O 


co ^t co oo r^ 




ooQmOcoosOawo 


rt '53 _, 


to co r^ ts co 


in 


r^OOooNinooot^ 


CQ ^ 








N M M MM 




V 




















•o 








c 


13 








.0 










a 

"5 


e . „ 








■js 








(0 
0) 


•o 








Q 












C O O 00 


O rf oo O O 


rfvO O OvO OO O 0>00 


c 
& 








M . 


O^ in to 


co in in in O* 


on 


O^O N O* O O^OOsint^i 


1/5 


^ HH_ M „ MMM MMM N MMMMMMMM 




«*» >-f* 


N O N O N 


-H* 


H« «t# 


•a 
c 


N Tf tO 


N 


WNWCINWWtNMM 


10 

| X X X 








■5 S. 


X X X X X X 


^g 








H* 




6*0 


H» 




O n- CO 


^ ^ O CO Tfr 


vO 


^•O ^ t * t * t « CO 




00 M XT) 


O O O m N 


co 


inO r^co ON coino^O 


rh in \n 


00 O M M M 




MMMMMNNNTj-in 









TRANSVERSE STRENGTH OF TIMBER. 



685 







usive, 
Oak 
r part 
ed at 
tested 
6. 


:ed Oak 
traight- 
ee from 








ij a si .- ~ 








.5 X! ci u O o* 


U £ 








_ > - a, 00 en 




1 

3 
& 




£* - V •- M > 






T * * 

. rt .2 § ° £° 

« S J3 Ei i* - 

2 o3 ^ £ t3 


go g 






2 

.2 
2 


S « « „; C 

« 1 .| § 5 

. $ rt 5 "-• 






O 


H 


* 8 














° £ 



t^ 


8888888R 

O^inOcNO m CO ^~ 


8 8 8 8 8 8 8 

CO en Tt O O «N CO 






D "0 


CO vO 


H 


8 5 


en rf M (N t>r^u-)vO 


in O en rf m en 


M 


Is 


«^-vo ■rt-r^.inr^o T 


en m en co vo u-> 


en 


00 m 


M •"^■M OO t^^O 


tN 0>00 M CO O 




s 


H 






H 


s« 










«M 










° s 


M O 


iNNMini-ienTj-vO 


O O tN O CO M t^ 


en 


J2 s 


VO M 


OcoNcoiitHcno 


in vO en ^t O 00 O 


vO 


O in 


en *i- 00 Tf t)- in 00 t^ 


CO N CO O CO <N M 


CO 


a a 


rf O 


v© in m in <t m tN ir> 


co «o •t O O r> 


in 


7? 3 

1* 










do .j- . 


M d 


<NTrin<NMM^M 


en O 00 "t O 00 » 




a j3 co 


en en 


« r-» r» 1^ O r^ r>. en 


i-O en «n 00 00 




'Chfl-fi 


C< O 


m rt« en in « N 


O m m o\ r>. i>. 




Breal 

Wei; 

inl 


Om O* O^ ^ 00 «N m 


in in rf ^ vO 






M M M 


M M M M M CN <N 






tu 


















. 


T3 








| 


•O 








<3 

.9* 

"C 


1 m 






C 


rt 






■0 

2 


cd 








3 






• 


a 

a 


.a 00 


OOcorfcocoOO 


CO vO vO O O 






CO O O 1 O^ N O 1 O O 


f«. in en r<» en 








H* H~ H» ^H ^H H°e h|h 


ic|co W H H* W H '-'!» »-*o 
M tN tN N O «N 0» 




a . 

rt J3 


ClMMlNMlMMM 




s M 






5 S. 

1° 


.5 xxxxxxxxxxxxxxxxx 

— fleet to .to to 

H» t-tao - «|« H» "rt H» "H "M t*H» 






TJ- ^t 


enen^enen'T'^Tj- 


^j- in in in in in *$• 




-i 


*-< N 


coco om <n en '<i- m 


O O O w ■* in O 






in m 


m <0 t*» f» r* »*• t^ 


r~> in 'O vO O vO O 

^1 M rj f^ Wt <*J M 








^ V« Vf VI *^ vl V* 





686 



APPLIED MECHANICS. 



WHITE-PINE BEAMS. 









Breaking 








No. of 


Width and 


Span. 


Centre 


Modulus of 


Modulus of 


Remarks. 


Test. 


Depth. 


Load, in 
lbs. 


Rupture. 


Elasticity. 




inches. 


ft. in. 










94 


3 X Hi 


15 8 


5088 


3613 


924252 


( Pattern stock. 


95 


3 X 13 


14 


12588 


7251 


1280832 


•< Clear piece. 
( Seasoned 3 yrs. 


96 


3 X 13 


16 6 


9088 


5324 


IO72889 




97 


3 X H 


15 8 


6088 


4729 


978256 




98 


2| X 9f 


16 


6088 


6415 


1234880 




99 


2| X 13 


15 6 


5988 


3438 


IO2039O 




100 


3 X 9* 


16 


4288 


4330 


II65937 




102 


3 X io| 


15 6 


4790 


3855 


99919O 




103 


3 X 11 


16 6 


6588 


5390 


1242649 




104 


3 X Hi 


15 6 


5088 


3739 


931760 




128 


6 X 12 


19 10 


12922 


5340 


I380660 




129 


6 X 12 


19 08 


15060 


6170 


15650OO 




130 


6 X i2£ 


19 08 


12340 


4954 


I222I00 




131 


6 X 12 


19 10 


13023 


538o 


I3O79O0 




132 


6 X 12 


19 10 


6231 


2575 


IIO35OO 




133 


6 X 12 


19 08 


12912 


5290 


I297OOO 




134 


6 X 12 


20 00 


II254 


4689 


1345700 




137 


6 T * X 12^ 


19 08 


13650 


5478 


I3677OO 




138 


6 X 12 


19 09 


14010 


5765 


I 247 I OO 


v 


140 


6 X 12 


19 09 


9761 


4016 


I IO560O 




244 


4AXI2H 


16 00 


10179 


4300 


94860O 




245 


4* X 12* 


15 6 


12984 


5620 


I27IOOO 




247 


4tV X 10$ 


15 8 


6770 


3638 


IIII9OO 




248 


4i X 12* 


15 8 


7790 


3549 


IO573OO 




279 


4 X 12* 


15 6 


9085 


4222 


IO84OOO 




280 


4 X 12* 


15 6 


7575 


35"2i 


8543OO 




281 


3f X 12^ 


15 


12070 


5547 


I28IOOO 




282 


3H X 12A 


15 


8660 


3998 


IO53OOO 




283 


3ff X I2± 


15 


7182 


3285 


9703OO 




284' 


4 X 12* 


15 


5685 


2456 


873300 




285 


3tt X 12A 


15 6 


6385 


3031 


9OI5OO 




286 


3| X i2 T V 


15 6 


11791 


5449 


I2587OO 




287 


4 X 12& 


15 6 


8265 


3802 


IO495OO 




288 


3* X I2 T % 


15 6 


11272 


5353 


1295382 




289 


31 X 12^ 


15 6 


5671 


2806 


8253OO 




296 


4 X 12& 


15 6 


557i 


2563 


727200 




315 


3l X 12 
Average valu 


16 
;s 


7165 


3820 


H584H 




4451 


I I 2200O 



It would seem to the writer that about the same modulus 
of rupture should be used for white pine as for spruce. 



TRANSVERSE STRENGTH OF TIMBER. 



687 



KILN-DRIED WESTERN WHITE PINE. 













Modulus of 


Modulus of 




No. of 


Depth and 
Width. 




Manner 


Breaking 


Rupture 


Elasticity 




Test. 


Span. 


Loading. 


Load (lbs). 


(lbs. per 
Sq. In.). 


(lbs. per 
Sq. In.). 


Remarks. 




inches. 


ft. in. 












206 


2 T V X i3t 


15 05 


Centre 


9325 


70I4 


15050OO 




207 


2 X I2| 


13 OO 




8814 


6432 


I I9300O 




208 


2 T V X I2fk 


15 OO 




3420 


2836 


752300 




210 


2 T V X I If 


15 OO 




4518 


4284 


I 24 I 400 




212 


2 T V X 13 


15 06 




6120 


4898 


IO99300 




213 


2AXI3H 


15 06 








127630O 




214 


2 X 12^ 


15 06 




7420 


6430 


I I66000 




237 


«A X 12H 


15 04 




8225 

Mean = 


6478 


I 23 I OOO 




5482 


H83037 



HEMLOCK. 



H 
"o 

0" 


Depth and 
Width. 


Span. 


Manner 

of 
Loading. 


Breaking 
Load (lbs.). 


Modulus 

of 
Rupture 
(lbs. per 
Sq. In.). 


Modulus of 
Elasticity 
(lbs. per 
Sq. In.). 


Remarks. 


inches. 


ft. in. 












154 


3i Xnf 


15 OO 


Centre 


6449 


3965 


870960 


Nos. 154-160 are 


155 


3tV X 10 


14 08 


<< 


4648 


4007 


971710 


Eastern hemlock 


r 5 6 


af X 9* 


12 09 




4425 


3716 


750400 


year. The re- 


157 


3t\ X ioi 


II IO 


" 


3223 


2381 


7709OO 


mainder of the 


158 
159 


3-A X loft 
3 X 9f 


14 OO 
13 06 


t( 


4137 
2939 


3151 

2570 


735800 
833000 


hemlock tests are 
from hemlock cut 
in Vermont June, 


160 


2| X ui 


12 08 




9433 


5911 


1086600 


1886; first growth 


T77 


4i X 12 


15 08 


tt 


9605 


4560 


1081100 


grown on high 


178 


4 X 12 


17 OO 


" 


5502 


2923 


821990 


ground. Sawed 
Sept. 25, 1886. Re- 
ceived at Institute 


179 


4i x nH 


15 08 




3192 


1531 


688960 


180 


4 X I2i 


15 06 


tt 


4584 


2175 


926560 


Nov. 15, 1886. Test- 


181 4 X"tt 


15 08 


«• 


5133 


2539 


758390 


ed Dec. 2, 1886, to 
March 9, 1887. 


182 4 X 12 


15 06 




1 1073 


5363 


1296600 


183 


4 X 12 


15 06 


" 


13274 


6499 


1269800 




184 


3* X n| 


15 08 


tt 


9964 


5142 


IO75600 




185 


3i X nil 
3ft X nif 


17 02 


'* 


3679 


2059 


412670 




186 


15 08 




12488 
Mean = 


6535 


T327200 




3825 


922250 


114 


7i X ioi 


20 04 


Centre 


19244 


8243 


20III88 


Yellow birch.N.H. 


120 


7* X ioi 


19 06 


" 


16150 


7627 


1583201 


" " " 


127 


3 X 10 


13 06 


" 


14965 


I2I22 




1 N. H. ash, sea- 
*) soned 2 yrs. 



688 



A P PLIED MB: CHA HICS. 



TIME TESTS. 



The following is a record of the time tests made at the 
Institute, and at the close will be found a statement in regard 
to the proper value of modulus of elasticity for use in com- 
puting deflections. 

TIME TEST NO. I. 

Spruce from Maine, received at Institute October 30, 1885. 
All the beams when received were green lumber, except F, 
which was seasoned on the wharf about six months. Beams 
A, B, C, D, E, and F were seasoned under a centre load in 
the laboratory in steam heat from November 10, 1885, to May 
8, 1886. Beams G, H, I were seasoned in the same room, 
without load ; span = 20 feet for all the beams under load. 



Beam. 


A (164) 


B (163) 


C (162) 


D (161) 


E (169) 


! 

F (168) 


Description of lumber, . 


clear 


knotty 


knotty 


clear 


clear 


clear 


Dimensions] IX- 1 .; 


4" x 12" 
3i"xni" 


4" x 12" 

4" XI If" 


4" x 12" 

3! " X I if" 


4" x 12" 
3f"xn|" 


6" x 12" 
5|"xii!" 


6" x 12" 

5t"*«i" 


f with'wt.of 














Max. fibre beam, . 


1078 


1076 


1074 


1070 


"33 


1136 


stress, lbs. -| without 














per sq. in. wt. of 














1. beam, . 


1003 


1003 


1003 


1003 


1057 


105 1 


Deflection, load first ap- 














plied, 


o".S488 


o".6 3 55 


o".76 7 5 


o".6io8 


o". 7 i88 


©"•5454 


Deflection at end of 














test, 


i".oi6 4 


1" -47°7 


i"-5i54 


1-3734 


1.2667 


0.3151 


E (immediate), . . . 


1462200 


1262900' 


1045700 


1 3 14000 


1 1 75000 


1540000 


E (final), 


789000 


546000 


529000 


584000 


667000 


2666000 


iKP ™r \ W^hoUt 
in S 'jn P 1 Wt. Of 

S( l- m - { beam, . 


6574 
6500 


7260 
7187 


5007 
4937 


6066 
6000 


6779 
6708 


8574 
8500 


Weight per cu. ft. at 














beginning, lbs., . . 


35>5 


34-9 


33-°" 


31-9 


34- 1 


35-4 


Weight per cu. ft. at 














end, lbs., 


27.1 


28.x 


25-5 


25 7 


28.2 


29.6 


Date of testing, . . . 


May 11, '86 


May 11, '86 


May 10, '86 


May 10, '86 


May 14, '86 


May 13, '86 


E (after seasoning), lbs. 
per sq. in. of final 


























section, 


1866900 


1509500 


1367300 


1625500 - 


i7374oo 


X718300 



Average E (immediate) = 1300000. 
Average E (final) = 963500. 

Average modulus of rupture beams under load a 6710. 

All quantities in the above table except the last were cal c u l at ed by oefagr the arigina* 
section. 



TRANSVERSE STRENGTH OF TIMBER, 



Beam. 



G (165) 



Description of lumber, 

_ . . 1 original 

D,mens,onS 1 final, 

Modulus of rupture, j with wt. of beam, . . 

lbs. per sq. in. 1 without wt. of beam, . 

Weight per cu. ft. at beginning, lbs., . . . 

Weight per cu. ft. at end, lbs., 

Date of testing, 

E (after seasoning), lbs. per sq. in. of final section 



clear 

4" x 12" 

3f"xnf" 

6525 

27-3 

May 11, '86 

1603700 



H (166) 



clear 

4 "XI2" 

3f"xia" 

7187 

27 

May 12, '86 
1748900 



I (167) 



knotty 

4"* 12' 
3013 
27.2 

May 12, ? 8f 

1457000 



Average E (final section), beams without load, 
Average modulus of rupture (original section). 



1603200 

6508 



TIMS TEST FOR SHORT PERIODS OF TIME. 











Total in- 






No. 


Depth and 
Width. 


Span. 


Original 
Modulus of 


creased 
Deflec- 


Modulus 
of 


Description of Test. 








Elasticity. 


tion. 


Rupture. 


• 




inches. 


ft. in. 




inches. 


lbs. 




108 


6X 12 


17 04 


1269670 


.1241 


5066 


Spruce-beam load equally dis- 
tributed at 12 points. The 
beam was subjected to a 
load of 5031 lbs. for 898 hrs. 


148 


4t 3 b X i ill 


17 04 


1211800 


•3765 


4668 


Yellow-pine beam, load dis- 
tributed equally over 12 
points. The beam was sub- 
jected to a load of 6355 lbs. 
for 29 days. 



TIME TEST NO. 2- 



Spruce beams cut in Maine in the spring of 1886. Received 
at Institute September 13, 1886. Beams A, B, C, D, E, and 
F were seasoned under a centre load, in the laboratory, in 
steam heat, from September 15, 1886, to April 2, 1887 ( 2 oo 
days). Beams G, H, I were seasoned in the same room, but 
were not loaded. All the beams were without load between 



690 



GPPL IED ME CHA NICS. 



April 2 and date of testing. Span == 18' 00" for all beams 
under load. 



Beam. 


A (194) 


B (193) 


C (191) 


D(igo) 


B(i9S) 


F(i93) 


Description of lum- 




clear j 




j- knotty-} 


straight. 


straight- 


ber, 


clear 


grained 


grained, 
some knots 


grained, 
some knots 




Dimen- j original, . 
sions 1 final, . . 


4 "XI2" 


4"*i*iV 


3*" XII*" 


4&"*«f" 


6" XI2&" 


6i" xm" 


3ir*»H" 


3H"*»H" 


3f"*»A" 


4*" xti||" 


5*"xnf" 


5&"*«H" 




with 
















wt.of 














Max. fibre 


beam, 


1020 


tosi 


1136 


994 


1095 


»93 


stress, lbs. 


with- 














per sq. in. 


out 
wt. of 
















, beam, 


966 


956 


1073 


943 


1037 


104s 


Deflection,load first 














applied, .... 


©"•3707 


o".4*53 


o"-5335 


o".4794 


o".30»4 


•"•5747 


Deflection at end of 














test, 


o".oi8i 


o".8447 


x".«336 
1333000 


«".©453 

1288000 


t".oiTO 
13s 7000 


J^.3030 

1173000 


E (immediate), . . 


1689000 


1449000 


E (final), . . . . 


682000 


730000 


577000 


591000 


655000 


448000 


„ « . f with 
Modulus | , 














, 1 wt. of 
of rup- 1 . 

H beam, 

ture, lbs. ■{ . . ' 

| without 














10283 


7838 


3»43 


5038 


7448 


5«tf 


persq ' wt.of 














in. I . 

1 beam, 


10329 


7783 


3183 


4987 


739« 


$064 


Weight per cu. ft. 














at beginning, lbs., 


31.8 


3»-8 


35-4 


30.0 


34-« 


30.5 


Weight per cu. ft. 














at end, lbs., . . . 


28.6 


s6.t 


*9.« 


37.© 


37.7 


36.3 


Date of testing, 


Apr. 25, *87 


Apr. 19, *8 7 


Apr. 5, *87 


Apr. 4, fy 


Apr. 37, ^7 


Apr.8,*87 


E (after seasoning), 














lbs. per sq. in. of 














final section, . . 


2125500 


1852300 


1731600 


X517800 


1662100 


S394MO 

. — 



Average B (immediate) ■« 1376500. 
Average E (final) ■» 614000. 

Average modulus of rupture beams 



TRANSVERSE STRENGTH OF TIMBER 



69I 



Dimensions ■ 



Description of lumber, 

original, 

final 

Modulus of rupture, j with wt. of beam, . . . 
lbs. per sq. in. 1 without wt. of beam, . . 

Weight per cu. ft. at beginning, lbs., 

Wt. per cu. ft at end, lbs., 

Date of testing, 

E (after seasoning), lbs. per sq. in. of final section, 



G (188) 



knotty 

4 "XI2" 

3*"xn|" 

5218 
5167 
20.5 
24.9 
Mch. 21, '87 
1355300 



H (187) 



clear 
4"xi 2 " 

3*S"*»*" 

8667 

8598 

35.8 

29-3 

Apr. 1, ^7 

1914500 



I (189) 



knotty 

4 "XI 2 " 

3i"x"&' ; 
4796 

475« 

27.3 

24.8 

Mch. 16, *87 

1573600 



Average E (final section), beams without load = 1614500. 
Average modulus of rupture (original section) =« 6227. 



TIME TEST NO. 3. 

Yellow-pine beams from Georgia, cut in season of 1886. 
Received at Institute September 13, 1887. The lumber was 
purchased in sticks of double length, and cut in two for testing. 
The numbers indicate the stick, and the letter " B " the butt, 
and " T " the top end of the same. Beams 1 T, 2 B, 3 B, 4 T, 
5 B, 5 T were seasoned under load, the remainder being sea- 
soned without load. Span = i&. 



692 



APPLIED MECHANICS. 



Beam. 


1 iT(» 3 8) 


a B (S39) 


3 B (240) 


4T(a4i) 


5 B (34a) 


5T(«43) 


Description of J 
lumber, . .| 


clear, 


clear, 


clear, 


clear, 


clear, 


clear. 


cross* 


some dry 


straight- 


cross- 


straight- 


straight- 


grained 


rot 


grained 


grained 


grained 


grained 


Dimen- ( original 
sions t final, . 


4lV'*"&" 


4H"*"&" 


4l"xi2 3 y 


3il"*ia&" 


6&"xn§r 


6 3 V / xi2i" 


4" *»*i" 


41s"*"*" 


3tf"*ii§i" 


3W'*"W 


6" x ii/ b " 


6i"xii«}" 




' with 














Max. 


wt. of 














fibre 


beam, 


1358 


»S9 


1381 


«435 


1626 


1509 


stress, • 


with- 














lbs. per 


out 














sq. in. 


wt. of 
















beam, 


1283 


"93 


130a 


1364 


I53« 


1426 


Deflection, load 














first applied, . 


o".53to 


o".s5«8 


o".47oo 


©".5618 


o".443« 


o"-4836 


Deflection at end 














of test, . . . 


©".9817 


t".»387 


c^.9434 


l".45*S 


©".8364 


o".8 44 9 


E (immediate), . 


1536000 


1 351000 


1763000 


1545000 


2290000 


1889000 


E (final), . . . 


832000 


602000] 


879000 


596000 


1213000 


108 1000 




' with 














Modu- 
lus of 
rup- 


wt. of 
beam, 
with- 


6290 


6334 


9545 


6570 


10630 


xoooo 


ture, 
















lbs. per 


out 
wt. of 














sq. in. 


beam, 


<Jat3 


6163 


94«t 


C495 


10540 


9984 


Weight per cu. ft. 














at beginning, 














lbs 


456 


40.6 


48.S 


43«* 


50.9 


45-8 


Weight per cu. ft. 














at end, lbs. . . 


38.4 


36.8 


43-3 


39.0 


45-8 


41.4 


Date of testing, . 


May as, *88 


May 28, '88 


May 28, '88 


May 28, *88 


May 31, '88 


June 1, '88 


E (after season- 














ing), lbs. per 














sq. in. of final 














section, . . . 


1898000 


1651000 


3240000 


1803700 


3501400 


3103400 

i 



Average E (immediate) ■= 1729000. 
Average E (final) » 867000. 

Average modulus of rupture beams under load m Stsz. 



TRANSVERSE STRENGTH OF TIMBER. 



693 



Description of lumber, 
Dimensions] ^^ 



Modulus of rup- 
ture, lbs. per 
sq. in. 



' with wt . of 
beam, . . . 
without wtc of 
beam, . . , 
Weight per cu. ft. at beginning 

lbs., 

Weight per cu. ft. at end, lbs., 

Date of testing, 

E (after seasoning), lbs. per sq 
in. of final section, .... 



B(m9) 



1 clear, 
■< straight- 
' grained 

4*"XI2&" 

4*"xnJ" 

9417 

9330 

46.1 
41.4 

March 28, 'J 

2141400 



»T<«3a) 



dry rot 

4A"x»f 

4&"x«A" 

3069 



April 6, '88 
1393000 



3T(^3) 



unsound 

4&"xia" 
4"««ir' 

5838 
5754 



April 9, '88 

1951000 



4BW) 



clear, 

straight* 

grained 

3§i"xi2*' 

3H"xia" 

9960 



March 26, '88 
1895600 



Average B (final section), beams without load = 1820200, 
Average modulus of rupture (original section) -» 7072. 



TIME TEST NO. 4. 

Spruce from Maine. Green lumber. Received at Insti- 
tute September u, 1888. Beams A, B, C, D, J, K were sea- 
soned under load in the laboratory in steam heat from Sep- 
tember 13, 1888, to March 25, 1889 (194 days). Beams E, F, 
G, and H were seasoned in the same room, but not loaded, 
All the beams were without load from March 25, 1889, to the 
date of testing. Span = 18' for the beams while under load. 



694 



A PPL IF. D MR CHA NICS. 



Beam. 


A («73> 


B(2 74 ) 


C (276) 


D (278) 


J (277) 


K(3 75 ) 


Description of lum- 








1 knotty, 


clear, 


clear. 


ber, 


ave. stock 


ave. stock 




•J straight- 
' grained 


straight* 
grained 


straight- 
grained 


Dimen- j original, . 
sions \ final, . . 


4*"*nH" 


4*"««H" 


4 A"*«i" 


4i"xji|" 


5H"*«lf" 


5f"xi2" 


4 "xni" 


3tt"x«A" 


4 "xxif" 


3 H"*»f s " 


5iV'*«i" 


5f"*»&" 




• with 














Max. fibre 


wt. of 














stress, 


beam, 


1423 


«44» 


1444 


1436 


1750 


«7°5 


lbs. per 


without 














sq. in. 


wt. of 
















. beam, 


1364 


«385 


1387 


1379 


1698 


1645 


Deflection, load first 














applied, .... 


o".6699 


a", 7716 


o".66x8 


o"-7379 


o".8446 


©".7439 


Deflection at end of 














test, ..... 


i".3023 
1326000 


*".8 3 75 
1x69000 


i".53°9 
1387000 


l".698c 
1224000 


i"-7359 

1308000] 


l". 4 28 4 

143x000 


E (immediate), 


E (final), . . . . 


682000 


491000 


597ooo 


532000; 


637000 


745ooo 


Modulus 
of rup- 
ture, lbs. 
per sq. 


• with 
wt. of 
beam, 
without 

! wt. of 


7545 


7211 


5597: 


6622 


6484 


7099 


1 beam, 


7487 


7»55 


5540 


6565 


6431 


7038 


Weight per cu. ft. at 














beginning, lbs., . 


34-3 


33-5 


33-o 


33-6 


31.5 


36 4 


Weight per cu. ft. at 














end, lbs., . . . . 


27-3 


25.4 


36.8 


36.9 


37-1 


28.6 


Date of testing, . . 


Mch. 26/89 


Apr. 3, '89 


Apr. 9, '89 


Apr. 11, *$g 


Apr. 10, *Sg 


Apr. 8, '89 


E (after seasoning), 














lbs. per sq. in. of 














final section, . . 


1631700 


1405500 


1747900 


1503200 


1576000 


1722600 



Average E (immediate) as 1307500. 
Average E (final), *■ 6x4000. 

Average modulus of rupture beams ander load m 5760. 



TRANSVERSE STRENGTH OF TIMBER. 



695 



Beam. 



£(270) 



F(2 7 I) 



G<273) 



H (2 5 J) 



Description of lumber, \ 

Dimensions \°£^* X 

^^rl^H^L^^^oi 

lbs. per sq. in. ^ beam 

Weight per cu. ft. at beginning, lbs., . . 
Weight per cu. ft. at end, lbs., .... 

Date of testing. 

E (after seasoning), lbs. per sq. in. of final 
section, 



nearly 
clear, 
straight- 
grained 
3*" x 12" 
3*"*nf" 

8293 
8237 

33-5 

27-3 

Mch. 13, 

1632500 



'89 



4"xnM' 

7615 

755* 

37-2 

27.6 

Mch. 19, '89 

1771000 



334 " X 12" 

slTxuft" 

8558 

8496 
36.8 
26.3 

Mch. 22, '89 
1770000 



knotty 

4&"xnJ" 

4ft" x iij" 

4732 

4676 

33 o 

27.7 

Nov. 6, '88 

1266500 



Average B (final section), beams without load 
Average modulus of rupture (original section) 



161000a 
7300. 



DEFLECTIONS WITH TIME. 

From the above it is plain that the deflection of a timber 
beam under a long-continued application of the load may be 
2 or more times that assumed when the load was first applied ; 
and in order to compute it by means of the ordinary deflection 
formulae, we should use for E not more than £ the value de- 
rived from quick tests. 

LONGITUDINAL SHEARING. 

Below are given tables showing the greatest intensity of 
the shear at the neutral axis of each beam at fracture as calcu- 
lated from the formula on page 675. 

For breaking shearing-strength per square inch, in the case 
of each wood, it seems to the author that it would be proper 
to use a value somewhere near the lowest of those given in 
the table of beams which gave way by longitudinal shearing. 

It will also be observed that these shearing-forces are less 
than those obtained from the experiments on direct shearing 
along the grain, made at the Watertown Arsenal ; and this is 
naturally to be expected, for the shearing in their case took 
place along a section that was perfectly sound, while in these 
cases it took place at the weakest point. 



6 9 6 



APPLIED MECHANICS. 



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TRANSVERSB STRENGTH OF TIMBER. 



697 



TABLE OF BEAMS WHICH GAVE WAY BY LONGITUDINAL 

SHEARING. 



Spruce 


Yellow Pine. 


Oak. 


White Pine. 




Intensity of 




Intensity of 


Intensity of 




Intensity of 


No. 


Shear, 


No. 


Shear, 


No. | Shear, 


No. 


Shear, 




Lbs. per Sq. In. 




Lbs. per Sq. In. 


I Lbs. per Sq. In. 




Lbs. per Sq. In. 


22 


202 


30 


273 


109 


152 


129 


155 


24 


190 


32 


242 


269 


379 


134 


119 


3* 


i54 


33 


i53 








233 


180 


35 


117 


50 


177 












36 


248 


147 


264 












46 


233 


4i9 


207 












90 


273 


429 


244 










... 


304 


130 


437 


289 










.... 


321 


233 


451 


151 






.... 






416 


215 


452 


240 










.... 


421 


199 
















423 


207 
















445 


i33 














.... 


450 


166 
















Average, 200. 


Average, 224. 


Average, 266. 


Average, 151. 



COMPRESSION OF TIMBER AT RIGHT ANGLES TO THE GRAIN. 

On page 698 will be found a table giving the averages of a 
series of tests of compression of timber at right angles to the 
grain. 

The pieces of timber tested were all about 13" long in the 
direction of the grain, the other two dimensions being as given 
in the table, the pressure being applied in the direction of the 
longer of these two dimensions. 

In the cases given in the tables a maximum load was found. 
Evidently, however, as the ratio of length in the direction of 
the pressure to least diameter decreases, we reach a point 
where no maximum load is found, but where continuous 
crushing goes on, the pressure continuously increasing. Thus,, 
in the case of spruce specimens 10" X 12", 4" X 6", 4" X 8", 
and 6" X 8", while a maximum load was sometimes found, at 
other times it was not. 



698 



APPLIED MECHANICS. 



COMPRESSION OF TIMBER AT RIGHT ANGLES TO THE 

GRAIN. 



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Average Compres* 
sive Strength per 
Square Inch. 


Spruce. ... • 

L 

Maple 

Hemlock 


lbs. 

2*X IO 

2 X 12 

3 XI2 

4 XI2 
6 XI2 

8 X12 

4 via 
4 xia 


60 

21 

2 3 
59 
26 
3i 

21 
90 


lbs. 

396 
252 
350 
397 
3i9 
35° 

1808 
33° 


f 

- 

Yellow pine.. -{ 

! 
1 

f 
Oak •( 


lbs. 

4x12 
6x12 
8x12 
8x 8 
4 x 6 

4x12 
4x12 
6x 12 

8 x 12 


60 
16 
18 
5 
17 

60 
51 
17 
21 


lbs. 

535 
402 

47i 
696 
566 

898* 
980* 

»35 
913 



* Two different lots. 



6°. Framing-yoints. —Another matter intimately connected 
with the strength of timber beams is the strength of the beam 
after it has been cut in some of the various ways commonly 
employed in framing. We are often told that a notch cut on 
top of a beam, or at the middle of its depth, or near the sup- 
port, does but little injury ; but the tests made, show the injury 
to be very large, amounting to a reduction of the strength of 
the beam to one-fourth or one-fifth of its original strength, 
with some of the most approved framing. 

The fact is that, in the case of timber, the shearing-strength 
along the grain is small, and that, in almost any case of notch- 
ing timber, as in a notched beam, or in a header, there is 
developed, in consequence of the cutting, a tendency to tear 



TRANSVERSE STRENGTH OF TIMBER. 



699 



the timber across the grain, and the resistance of timber to this 
Icind of stress is very small. Moreover, the injury due to 
notching, so far from being a small quantity, is very large. 
Hence it follows that almost any cutting does a great deal 
of injury ; and it is much better to avoid framing whenever 
it is possible, and use stirrup-irons instead. In these tests, 
only two of the most approved framing-joints have been 
tested; viz., the joint known as the " tusk-and-tenon," shown 
in Fig. 243, and used for framing the tail-beams of a floor into 
the headers, and the "double 
tenon and joint bolt," shown in 
Fig. 244, and used for framing 
the headers into the trim- 
mers. 




\ 13 



Fig. 243. 



Fig. 244. 



The arrangement is shown in plan in Fig. 245, where I and 
2 are the trimmers, 3 is the header, and 4, 5, and 6 are the 
tail-beams ; the latter being supported at 
one end on the header, and at the other 
on the wall, the header being supported 
by the trimmers, and the trimmers being 
supported on the walls at both ends. 

It is sometimes the practice to hang 
the header in stirrup irons, and this is an 
improvement ; but it is very seldom that 
the tail-beams are hung in stirrup irons, 
and these tests have shown the weakening 
already referred to, from the mortises cut 
in the header to admit the tail-beams. 

A spruce floor was first built and 
tested, the following being a partial ac- 
Fig *4 S . count of the test : — 

No. 52.^ — Section of a floor between the trimmers. Spruce : 
three tail-beams, 2 inches by 12 inches each, framed into a 31- 
inch by nf-inch header; header in turn framed into sections 



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> 4 


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V .5 


D 


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N 6 


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1 
1 


/ 


M < 



700 APPLIED MECHANICS, 

of the trimmers by double tenon and joint-bolt, cross-bridged in 
two places ; tail-beams framed by tusk-and-tenon joint, pinned, 
floored over and furred below ; load at centre, distributed be- 
tween the three tail-beams by bridging. 

Span = 16 feet ; weight of joist, flooring, etc., = 331 lbs. 

1 1238 lbs. = breaking-load. 

Joist on east side broke by splitting off at the tenon, bore 
7988 lbs. after. The load was then increased. Centre tail-beam 
broke by tension at 9988 lbs., on account of cross-grain in the 
lower fibres. A split also started at the lower tenon of the 
header, which at the time of breaking was rapidly increasing. 

Average modulus of rupture of the tail-beams, including 
their own weight, etc., = 3801 lbs. per square inch. 

Average modulus of elasticity of tail-beams = 1399141 lbs. 
per square inch. 

It is to be noticed, that the header already began to crack 
when the tail-beams broke, and hence that the floor could have 
borne but little more, even if the load had been uniformly dis- 
tributed : hence that, in this case, the breaking-strength of the 
floor would be determined by calculating the loads at the centre 
of the tail-beams, instead of accounting it as distributed ; in 
other words, the breaking-weight would be about one-half what 
we should get by considering the load as distributed on the 
tail-beams. 

YELLOW-PINE HEADERS. 

A number of tests of the strength of yellow-pine headers, 
and also of spruce headers, have been made in the Laboratory 
of Applied Mechanics of the Massachusetts Institute of Tech- 
nology, and the results will be given here. 

It will be seen from these tests that the first of these head- 
ers had for its breaking-weight 13 163 lbs., and the second 
11631, or in each case one-half the load on the floor. To 
institute a comparison, we may observe that, if a 6-inch by 
12-inch yellow-pine header 6 feet 8 inches long, with four tail- 



TRANSVERSE STRENGTH OF TIMBER, 



701 



beams 18 feet long, were to support a floor, the floor surface 
would be 96 square feet, giving 48 square feet to be supported 
by the header. This, if the floor were loaded with 100 lbs. per 
square foot, would bring upon the header 4800 lbs., or about 
one-half the breaking-weight of a header only 5 feet 4 inches 
long ; whereas, it would commonly be supposed, that, with such 
a construction for 100 lbs. per square foot of floor, we should 
have provided an unnecessarily large margin of safety. 

The special source of weakness in the header is, of course, 
the joint by which the tail-beams are attached to it, while the 
framing of the header into the trimmer causes great loss of 
strength in the trimmer. Nevertheless, even for the sake of 
the header, hanging it in stirrup-irons on the trimmer is better 
than framing. 

The fact, also, that a 6-inch by 12-inch yellow-pine beam 5 
feet 4 inches long bore 48000 lbs. centre load, equivalent to 
96000 distributed, without breaking, while the header broke at 
10916, shows what an enormous weakening is caused by cutting 
mortises, and how much strength would be gained by avoiding 
all framing, and using stirrup-irons to support the tail-beams in 
all cases where they cannot be supported on top of the header 
bearing the latter. 

TESTS OF YELLOW-PINE HEADERS. 




Fig. 246. 



The yellow-pine headers were all 6 inches wide by 12 
inches deep, and, in every case, the tail-beams were placed 
16 inches apart centre to centre, so that the length of the 



702 APPLIED MECHANICS. 

header between the trimmers, in any case, can be found by- 
multiplying sixteen inches by one more than the number of 
tail-beams. The headers were either framed into the trim- 
mers by a double tenon and joint-bolt, or else they were hung 
from them by stirrup-irons, the trimmers being supported upon 
jack-screws. The headers were mortised at the proper places 
(sixteen inches on centres) for twelve-inch yellow-pine tail- 
beams sometimes ten feet length between headers, sometimes 
six, but more often three feet between the headers. 

The load was applied at the centre of the tail-beams, and 
divided equally among them by iron bars and knife edges. 

The longer tail-beams were cross-bridged by 2" X 3" 
spruce bridging, but the shorter ones were not bridged. 

The mortises in the headers were, in every case, those suit- 
able for a tusk and tenon joint, the tail-beams being in most 
cases three inches wide. 

The table giving the results of the tests of the yellow-pine 
headers will be found on page 703. 

In order to form an adequate conception of the amount of 
the loss of strength of one of these yellow-pine headers carry- 
ing three tail-beams, assume the same beam with no notches, 
but with the load equally divided at three points sixteen inches 
on centres, compute the breaking strength of such a beam, 
and compare with it the strength of any one of the headers 
tested which carried three tail-beams. Use for modulus of 
rupture 5000 pounds per square inch. 

Performing the calculation, we easily obtain for the break- 
ing strength of the six-inch by twelve-inch yellow-pine beam 
without the notches 67500 pounds. The average of the break- 
ing strength of the four headers carrying three tail-beams is 
13127 pounds, which is only 0.194 of 67500 pounds; hence 
the notches for the tusk and tenon joints where the tail-beams 
were attached to the headers caused a loss of about eighty per 
cent in the strength of the latter. 



TRANSVERSE STRENGTH OF TIMBER. 



703 



The following table shows the results of the tests of the 
yellow-pine headers. 

YELLOW-PINE HEADERS. 



Number 


Width 


Test. 


(inches). 


89 


6 


I05 


6 


106 


6 


107 


6 


482 


6 


483 


6 


484 


6 


485 


6 


486 


6 


487 


6 


490 


6 


492 


6 


493 


6 


500 


6 



Depth 
(inches). 



Length be- 
tween Trim- 
mers (inches). 



64 

64 

64 

64 

80 

80 

96 

96 

96 

112 

128 

80 

112 

96 



Number 

of 

Tail Beams. 



Breaking- Load 

of each Header. 

Pounds. 



I3163 
II631 

I5I3I 
I2581 
15190 
2019O 
19870 
16045 
16925 
I86IO 
13905 
12795 
I30IO 
16370 



DETAILS OF THE TESTS. 

No. 89. — The headers were framed into the trimmers by 
double tenons and joint-bolts. 

The tail-beams were each 3 inches by 12 inches, and at first 
were 10 feet long, the result being that, at a total load of 
24866 pounds, i.e., 12433 pounds on each header, one of the 
tail-beams broke under the tenon by splitting, while the head- 
ers* were left intact. These tail-beams were then removed 
and new ones were supplied each 3 inches by 12 inches as be- 
fore, but only 80 inches long, and the test was repeated, re- 
sulting in the breakage of one of the headers by splitting- 
through the middle, following the line of mortises. 

No. 105. — The headers were framed into the trimmers by 
double tenons and joint-bolts. The tail-beams were each 3, 
inches by 12 inches, and 72 inches long. 



7° A APPLIED MECHANICS. 

One of the headers failed through the line of mortises. 

No. 106. — The headers were supported on the trimmers by 
means of stirrup-irons. 

The tail-beams were 3 inches by 12 inches, and 72 inches 
long. 

At a total load of 26,662 pounds, i.e., a load on each header 
°f 13331 pounds, one of the tail-beams split below the line 
of mortises. The total load was then increased to 30,262 
pounds, i.e., 15 131 pounds on each header, when one of the 
stirrup-irons broke, but simultaneously with this the header 
failed. 

No. 107. — The unbroken headers of Nos. 105 and 106 
were used for this test, supported on the trimmers in stirrup- 
irons. At the maximum load one of the headers failed sud- 
denly. 

Of the remaining headers the last three were supported 
on the trimmers in stirrup-irons, the other seven being framed 
into the trimmers. 

No. 482. — One of the headers split, starting at the lower 
tenon at one of the trimmers. 

No. 483. — One of these headers was the unbroken one of 
No. 482. This header broke in the same way as in the case of 
No. 482. 

No. 484. — One of the headers split. 

No. 485. — One of the headers split. 

No. 486. — One of the headers was the broken one of No. 
483, and this one failed by splitting. 

No. 487. — One of the headers split. 

No. 490. — The splitting of one of the headers was followed 
almost immediately by that of the other. 

No. 492. — The failure of one header was soon followed by 
that of the other. 

No. 493. — One of the headers failed, splitting very much. 

No. 500. — One of the headers split. 



TRANSVERSE STRENGTH OF TIMBER. 



70S 



TESTS OF SPRUCE HEADERS. 

The general dimensions of the floors tested are shown in 
the figure. In all these tests the tail- 
beams are joined to the headers by tusk 
and tenon joints. The load was dis- 
tributed equally at the centres of the 
three tail-beams. 

The following table shows the results 
of these tests. 

SPRUCE HEADERS (Figure, page 705;. 







1 r 


1 










5 *: 3 

(3) 




t , 




(Sft 


4 


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"to 


it 








1 f 


i "1 


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1 1 


J 



Number 


Width 
(inches). 


Depth 
(inches). 


Length be- 


Number 


Breaking Load 


of 


tween Trim- 


of 


of each Header. 


Test. 


mers (inches). 


Tail Beams. 


Pounds. 


141 


4 


12 


64 


3 


9020 


142 


4 


12 


64 


3 


9920 


143 


4 


12 


64 


3 


9415 


170 (a) 


3f 


12 


64 


3 


6917 


170 {6) 


3f 


12 


64 


3 


7417 


170 {c) 


3! 


12 


64 


3 


9417 




SPRU 


CE HEADERS (Figure, j 


age 706). 




217 {A) 


4 


12 


64 


3 


I OO00 


217 (B) 


4 


12 


64 


3 


15850 


217(C) 


4 


12 


64 


3 


10450 



DETAILS OF THE TESTS. 

No. 141. — Headers were framed into trimmers by double 
tenons and joint-bolts. Tail-beams were made of spruce. One 
of the headers broke by tension. 

No. 142. — Headers were supported on trimmers by stirrup- 
irons. Spruce tail-beams were used at first, but they broke, 
leaving the headers intact. Then yellow-pine tail-beams were 
used. One of the headers failed by splitting along the' middle. 



<f-13 15 K 15 13-j - 

ja S EL 



706 APPLIED MECHANICS. 

No. 143. — These headers were the unbroken ones of Nos. 
141 and 142 ; the first one framed into the trimmer, the other 
hung from the trimmer by stirrup-irons. The header framed 
into the trimmer failed by splitting. 

No. 170 (a). — Headers joined to trimmers by double tenons 
and joint-bolts. One of the headers split. 

No. 170 (b). — Unbroken header of previous test used for 
one of the headers. One header split. 

No. 170 (c). — The unbroken header of the previous test 
used for one of the headers. One header split. 

No. 217. Dimensions of floor altered as indicated in the 
figure. 

217A. — Headers framed into trimmers. 
One header split. 

217B. — Headers supported upon trim- 
ly mers by stirrup-irons. One header split. 
J 217C. — Unbroken header of 217A used 
for one of the headers. One header split. 

GENERAL REMARKS. 

The stresses brought into play by the load on a header 
are not only bending and shearing, but also a tension across 
the grain, and any method of figuring the load a header 
would bear without taking account of all three, and especially 
the latter, would not furnish correct results. 

Moreover the character of the results of tests of yellow- 
pine headers of different lengths, given on page 703, confirm 
this statement, for the strength with different lengths is not 
by any means universally proportional to its length, and, on 
the other hand, the breaking strengths do not increase with 
the length, as they would do if tension across the grain were 
the only stress and bending did not take place. 



TRANSVERSE STRENGTH OF TIMBER. 



707 



BAUSCHINGER'S TESTS. 

In the ninth Heft of the Mittheilungen are given the results 
of an experimental study which Bauschinger made of the 
strength of certain pine and spruce woods, in connection with 
their other properties, as specific-gravity, age, time of felling, 
etc. ; but special attention is given to the variation of strength 
and specific gravity, with the percentage of moisture which they 
contain, i.e., their condition of dryness. While he did a con- 
siderable amount of work upon the variation of the tensile 
strength with the percentage of moisture, the results are rather 
variable, and none of this work will be quoted here for the 
reasons given on page646, under Tension of Timber. He him- 
self came to the conclusion that more satisfactory results could 
be reached by experimenting upon compressive strength. 

The following tables give summaries of the tests which he 
made and reported in this ninth Heft upon compressive and 
transverse strength : 



COMPRESSIVE TESTS. 
Test-Pieces about 3J x 3^ Inches and 6 Inches Long. 



Timber. 


Place. 


Summer Felled. 


Winter Felled. 


Percent- 
age of 
Moisture. 


Compressive 

Strength. 

Mean Values. 

Lbs. per Sq. In. 


Percent- 
age of 
Moisture. 


Compressive 

Strength. 
Mean Values. 
Lbs. per Sq. In. 


Pine . . 
Spruce . 
Spruce . 
Spruce . 


Lichtenhof . 
Frankenhofen 
Regenhiitte . 
Schliersee . 


9 
20 

27 
20 


3997 
3449 
3328 
2304 


26 

17 
20 

T 9 


4537 
4452 
3997 
3200 



708 



APPLIED MECHANICS. 



TRANSVERSE TESTS. 



Timber. 



Place. 



Breadth and 
Depth. 
Inches. 



Per- 
centage 
of Mois- 
ture. 



Modulus of 

Elasticity. 

Lbs. per 

Sq. In. 



Modulus of 

Rupture. 

Lbs. per 

Sq. In. 



I. Summer Felled. Span, 8 ft. 2.43 ins. 



Pine 



Spruce 



Lichtenhof 



Frankenhofen 



Regenhiitte 



Schliersee 



6.70X 

7.19X 

6.66X 

7.06X 

6.72X 

7.32X 

7-77X 

8.10X 

7-52X 

8.01X 

7-88X 

8.36X 

10.17X 

10.25X 

10.47X 

10.73X 



6.81 


24 


7.17 


23 


6.72 


19 


7.17 


25 


6.76 


29 


7-25 


35 


7.80 


25 


8.10 


26 


7.64 


39 


8.07 


34 


7.83 


30 


8.26 


32 


10.24 


25 


10.34 


24 


10.47 


21 


10.67 


24 



1422300 
1536084 
1536084 
I 66409 I 
1706760 
1649868 
1464969 

1436523 
1578753 
1592976 

1635645 

1720983 

1016945 

960052 

I 109394 

1080948 



6002 

6685 
6742 

7453 
6372 

6344 
5703 
5405 
6016 
5476 
6173 
6016 
4025 
3840 
4281 
4281 



II. Winter Felled. Span, 8 ft. 2.43 ins. 



Pine . 



Spruce 



Lichtenhof . 



Frankenhofen 



Regenhiitte 



Schliersee 



6.58X 6.71 


33 


6.23X 6.30 


33 


7.I9X 7.34 


3i 


8.30X 8.17 


34 


7.07X 6.94 


31 


7.I9X 7.29 


28 


6.65X 6.69 


24 


6.64X 6.67 


24 


6.38X 6.39 


27 


7.02X 7-o8 


29 


6.66X 6.79 


30 


7.65X 7-66 


38 


10.98X12.66 


26 


10.83X13-18 


25 


n. 19X11. 23 


28 


ii.iiXii.io 


25 



1422300 
1664091 
1308516 

1479192 
1479192 

1436523 

1863213 
1820544 

1479192 

1536084 

1635645 
1635645 

1024056 

981387 

967164 

981387 



6813 
7609 
5348 
5903 

5959 
5903 
6969 
6813 
6230 
6329 
6443 
6358 
3641 
3755 
3670 
3570 



TRANSVERSE STRENGTH OF TIMBER. 709 

In Heft 16 of the Mittheilungen, Bauschinger gives an 
account of a series of tests of the crushing and transverse 
strength of the more important coniferous woods from the dif- 
ferent districts of Bavaria. 

In the case of the transverse tests the percentage of moisture 
was determined by experiment, and is recorded in the tables. 
Then sections were cut from the same pieces from which the 
beams were taken, and tested for crushing, one of them being 
rather wet; one had somewhere near 15 per cent of moisture, 
which Bauschinger considers to be about the average dryness 
of the air, and one was somewhat drier ; and in each case the 
percentage of moisture is determined and recorded. 

The results of the crushing tests are then plotted, and a 
curve drawn, from which he determines the crushing-strength 
with 15 per cent of moisture. A similar proceeding is adopted 
in regard to the specific-gravity. 

The 45 sections were each cut into five specimens about 3 
or 4 inches square, one of them containing the heart. 

These (which contain the heart) he omits from his curves 
and calculations, and plots only the results of the others. 

His results are given in the following table, a perusal of 
which will show that the moduli of rupture, and also the crush- 
ing-strengths, run somewhat higher than they do for woods of 
the same name in the tests made at the Massachusetts Institute 
of Technology. This, of course, maybe due to the woods that 
Bauschinger tested being stronger than American woods of the 
same name, but it is more probably due to the facts that, i°, the 
specimens he used were rather smaller, and, 2°, they were, on 
the whole, drier than the American woods tested at the Mas- 
sachusetts Institute of Technology. 



710 



APPLIED MECHANICS. 









? 


Mean Compressive 
Strength of 


















Transverse Test. 













Pieces not con- 




Span, 8 ft. 2.42 ins. 










•a 


taining Heart, 










Name. 


Place of 
Growth. 





11 
u 

>» . 

22 


reduced to 








a 
15% 


b 

Same 

percent 

Moisture 






•5 
'S 

s 

O 


Modu- 
lus of 


Modu- 
lus of 
Rup- 


08 






OS 


u 

0. 


Moisture. 


as in 






V 


Elas- 


i 

o 

d 






Lbs. per 
Sq. In. 


Trans- 
verse 
Test. 
Lbs. per 


5« 

u C 




ticity. 
Lbs. per 
Sq. In. 


ture. 

Lbs. 

per 

Sq.In. 


£ 






to 




Sq. In. 


CQHH 


QJ5 


Ph 






i 


Larch . . 


St. Zeno . . 


0.62 


6827 


6756 


5-78 


6.97 


15-5 


2076560 


10382 


2 


Larch 




'* 


. 


0.67 


7353 


6784 


5-85 


6.99 


17.2 


1692540 


10596 


3 


Pine 




" 




0.53 


6045 


6400 


5-78 


5-74 


13-5 


1834770 


9742 


4 


Spruce 




" 


. . 


0.43 


4978 


455i 


7-78, 


7.61 


16.6 


147919° 


7183 


5 


Pine 




" 


. . 


0.52 


5263 


5120 


7-75 


7.70 


15-8 


1649888 


7325 


6 


Spruce 




" 


. . 


0.45 


5405 


5334 


7.69 


9.21 


15-4 


1635650 


6756 


7 


" 




" 




0.48 


5547 


5831 


5-8o 


6.85 


14-3 


1592980 


7965 


8 


" 




" 




0.51 


5974 


5689 


6.87 


6.82 


16. 1 


1578750 


775i 


9 


Larch 




it 




0.59 


6685 


4978 


6.10 


6.91 


16.0 


1706760 


9245 


IO 


'* 




" 




0.58 


6n6« 


5974 


5-85 


6.91 


15-5 


1521860 


9743 


ii 


Ziirbe 




" 




0.41 


3200 


3271 


5-86 


7.00 


14.7 


846269 


519 1 


12 


Larch 




Karlstein 




0.61 


7965 


7752 


5-8 5 


7.16 


15.8 


2097898 


9529 


13 


Spruce 




" 


. 


0.54 


5974 


5334 


6.79 


7.78 


17.4 


1592980 


7965 


14 


Pine 








0.54 


5974 


6258 


5-6 5 


6-93 


14.4 


1905880 


10027 


15 


Spruce 




'* , 




0.46 


5405 


519 1 


6.78 


7.80 


16.3 


1777880 


8107 


16 


Larch 




ti 


. 


0.68 


7965 


7325 


7.61 


9-34 


17.9 


2039670 


8605 


17 


Spruce 




41 


. 


0.54 


6969 


6898 


5-79 


6-95 


i5-i 


2026780 


9387 


18 


'* 




" , 


. 


0.49 


6045 


5476 


5-86 


7.80 


18.3 


1578750 


6542 


i9 


" 




** 




0.52 


6187 


6045 


5-83 


6.87 


15.6 


1692540 


8036 


20 


Pine 




" 




0.52 


5831 


5903 


5-77 


7.71 


14.9 


1550307 


7823 


21 


Larch 




" 




0.61 


7040 


6258 


5- 81 


6-93 


17.6 


1905880 


10241 


22 


Spruce 




" , 




0.51 


6471 


6187 


5-76 


7-77 


15.8 


1564530 


8107 


23 






" , 


. 


0.39 


4267 


4054 


7.72 


9-31 


15-9 


1208960 


5334 


24 


Larch 




Freising 




0.61 


7823 


7538 


5-75 


6.94 


15 -9 


2062340 


9814 


25 


" 




" 




0.70 


8249 


7538 


5-84 


6.91 


16.9 


2176120 


9814 


26 


Spruce 




" 


. 


0.59 


6969 


5831 


6.87 


7.72 


18.4 


1977000 


8249 


27 


" , 




" 


. 


0.44 


5405 


5I9 1 


6.96 


7.62 


15-6 


1479190 


6685 


28 


Pine 




" 


. 


0.52 


6187 


5476 


6.88 


7-75 


18.4 


1550310 


6258 


29 


" 




11 




0.62 


7894 


7467 


6.86 


7.64 


!6. 9 


2062340 


9814 


30 


Spruce 




11 


. 


0.48 


5974 


5689 


6.91 


7-74 


16.7 


1607200 


7609 


31 


<i 




" 


. 


0.47 


5618 


4480 


6-93 


7.80 


18.6 


1763650 


6898 


32 


Pine 




11 


. 


0.53 


6116 


5689 


5-8 3 


6.86 


16.8 


1848990 


8818 


33 


" 




" 


. 


0.49 


4907 


455i 


5-8 3 


6.91 


17.0 


1223180 


5974 


34 


Spruce . 




" 


. 


o-59 


6827 


6400 


6.82 


7-74 


16.7 


1948550 


9103 


35 


" , 




" 




0.48 


5618 


5334 


6.96 


7.70 


16.4 


1 6783 10 


6471 


36 


" . 




" 




0.44 


5334 


5I9 1 


6.96 


7.71 


15.8 


1550310 


7325 


37 


" 




" 


. 


0.49 


6258 


5974 


7.00 


7.82 


16.4 


1806340 


7467 


38 


Larch 




it 


. 


0.58 


7325 


7112 


5-77 


6.91 


15-5 


1742320 


9387 


39 


" 




" 




0.65 


5903 


5263 


7.00 


6.96 


18.2 


1550310 


7752 


40 


Pine 




it 




0.49 


4764 


4267 


6.96 


7.81 


17.7 


1038280 


3485 


41 


Spruce 




k 




0.46 


6116 


5547 


6.88 


7.80 


17.0 


1280070 


5618 


42 


" 




Unterliezheim 


0.42 


4120 


4125 


6.85 


7.71 


20.7 


1464970 


6144 


43 


" 




" 


o.39 


4907 


455i 


6-95 


7.80 


18.8 


1251620 


5I9 1 


44 


White Pine 


tt 


o-33 


3414 


3058 


6.99 


7.83 


18. 1 


8 1073 1 


4125 


45 






0.32 


3200 


2631 


7.00 


7.84 


21.4 


725370 


3556 



TRANSVERSE STRENGTH OF TIMBER. 



711 



AVERAGE COMPRESSIVE STRENGTH OF WHOLE SECTION OF LOG. 
Pounds per Square Inch. 



Time of 
Felling. 


Pine from 
Lichtenhof. 


Spruce from 
Frankenhofen. 


Spruce from 
Regenhlitte. 


Spruce from 
Schliersee. 


i 


2 


1 


2 


1 


2 


1 


2 


Summer 
Winter 


7183 
6343 


5244 
6784 


6416 
6756 


4807 
56l8 


6287 
6343 


5319 

5348 


4570 

4779 


3143 

4238 



1. Tested 5 years after felling. 2. Tested 3 months after felling. 

OTHER FULL-SIZE TESTS. 

References to other full-size tests of timber are: 

i°. Tests of Pine Stringers, and Floor Beams, by Onward Bates, 
Trans. Am. Soc. C.E., 1890. 

2 . Tests of White Pine of Large Scantling, by Prof. H. T. Bovey, 
Trans. Canadian Soc. C.E., 1893. 

3 . The Strength of Canadian Douglas Firs, Red Pine, White 
Pine and Spruce (mostly full size). Trans. Canadian Soc. C.E., 1895. 

4 . The Proc. Fifth Annual Convention of the Assoc. R. R. Supts. 
contains, among others, the follqwing references : (#) Tests of 
Strength of State of Washington Timbers, Talbot Hart, and S. K. 
Smith ; (b) Tests of the Northwest and Pacific Coast Timbers, S. K. 
Smith and Thurston ; (c) Tests of California Redwoods, Soule ; 
(d) Old and new White Pine Stringers, Finley ; (e) Beams of 
Douglas Fir, Wing, 1895. 

5 . U. S. Dept. of Agriculture ; preliminary circular issued by 
the Bureau of Forestry, giving outline of future work, 1903. 

6°. U. S. Dept. of Agriculture ; Progress Report on the Strength 
of Structural Timbers, Bureau of Forestry, 1904. 

§ 240. Shearing of Timber along the Grain. — The shear- 
ing of timber almost takes place along, and not across, the grain; 
for it can be shown, that, wherever we have a tendency to shear 
on a certain plane, there is an equal tendency to shear on a plane 
at right angles to it. Hence if there is, at any point in a piece 
of wood, a tendency to shear across the grain, there must neces- 
sarily accompany it an equal tendency to shear it along the grain; 
and wherever (as is almost always the case) the resistance to the 
latter is less than the resistance to the former, the timber will give 
way in this manner, instead of across the grain. As to the shear- 
ing-strength per square inch, some values have been given in Ran- 



712 



APPLIED MECHANICS. 



kine's table; and the following table contains results obtained at 
the Watertown Arsenal, and recorded in Tests of Metals for 1881. 







Shearing- 






Shearing- 


Kind of Wood. 


Arsenal 


Strength 


Kind of Wood. 


Arsenal 


Strength 




No. 


per Square 
Inch. 




No. 


per Square 
Inch. 


Ash 


620 


600 


Oak (white) . . 


631 


752 




621 


592 


Pine (white) . . 


752 


324 




622 


458 




753 


267 




623 


700 




754 


352 


Birch (yellow) 


623 


563 




755 


366 




633 


815 


Pine (yellow) • . 


607 


399 




634 


672 




608 


3 l 7 




635 


612 




614 


409 


Maple (white) . . 


636 


647 




615 


4i5 




637 


537 




616 


409 




638 


367 




617 


364 




639 


43i 




618 


286 


Oak (red) . . . 


624 


775 




619 


330 




625 


743 


Spruce .... 


748 


253 




626 


999 




749 


374 




627 


726 




750 


347 


Oak (white) . . 


628 


966 




75 1 


316 




629 


803 


Whitewood . . 


609 


406 




630 


846 




610 


382 



§ 241. General Remarks. — A perusal of the tests on 
columns and on beams will show that one of the principal 
sources of weakness in timber is the presence of knots, and it 
will be noticed that the position of the fracture is in most 
cases determined by the knots. 

Sap-wood, season cracks, and decay are doubtless other 
sources of weakness. The tests, however, do not present such 
striking evidence of the deleterious effects of the first two as 
is the case with knots. In general, it may be said, however, 
that timber used in construction should be free, or nearly free, 
from sap-wood ; as an excessive amount of sap-wood renders it 
weak. 



GENERAL REMARKS. 713 

It will often be found to be a common opinion among lum- 
ber-dealers, that a piece of timber which contains the heart is 
not as good as one which is cut from the wood on one side of 
the heart. This is very often true ; as the timber which is sold 
in the market is very liable to have cracks at the heart, and 
also, if the tree has passed maturity, the heart is the place 
where decay is likely to begin. Nevertheless, the tests of 
beams would not, it seems to the author, bear out the conclu- 
sion that such pieces as contain the heart are always weaker 
than those that do not. 

Another matter that claims serious consideration is the 
effect of seasoning upon the strength of timber. This question 
can only be decided by tests on full-size pieces, as the small 
pieces season much more rapidly and uniformly than full-size 
pieces. 

In this regard, the observation should be made, that prac- 
tically our buildings and other constructions are built with 
green lumber; i.e., lumber which has been cut from three 
months to a year. Unless it can be shown that the seasoning 
which the lumber receives while in use imparts to it a greater 
strength, it will only be proper to consider its strength the same 
as that of green lumber. Not very much evidence has thus far 
been obtained upon this point ; but, such as it is, it will be 
noted here. 

i°. We have, on p. 653, the results of the tests of a lot of 
old mill columns ; and, while some of them did exhibit a greater 
strength than green ones, a perusal of this set of tests will 
convince the reader that it would not be safe to rely upon any 
greater strength in these columns than in green ones. More- 
over, these columns had been in a building heated by steam for 
a number of years, and during the seasoning process they had 
been subjected to the load they had to support. The writer 
has also observed some evidence of the same kind in con- 
nection with one of his time tests. 



714 APPLIED MECHANICS. 

2°. In the case of beams, we have, in Nos. 60 and 66, 
examples of beams which had been seasoning, unloaded, in a 
building heated by steam ; and in these cases there was a great 
gain in strength. Some yellow-pine beams exhibited a similar 
action. On the other hand, beams Nos. 18 and 19 had been 
seasoning on the wharf, in the open air, for about one year ; 
and while some yellow-pine beams which had seasoned without 
load, in the building, showed great strength, in other cases the 
increase was not so marked. 

In view of the fact that the above is practically all the evi- 
dence we have in the matter, it would seem to the writer, 
unless future experiments shall prove the contrary to be true, 
that we cannot rely, in our constructions, upon having any 
greater strength than that of the green lumber, and that the 
figures to be used should be those obtained by testing green 
lumber. 

§ 242. Building-Stones. — The three most important factors 
about a building-stone besides its beauty, are its durability, its 
strength, and the ease with which it can be quarried and 
worked. In order to be durable it must be able to withstand 
the deleterious influences of rain, wind, frost, fire, and of the 
acids that are found in the air especially in large cities, where 
the most common are carbonic acid and sulphur acids. 

The durability of a stone is probably its most important 
feature, and is, perhaps, the most difficult to test thoroughly. 
As a rule, the greater its hardness and the less its absorptive 
power for moisture, the more durable will it be. 

Tests of hardness are easily and frequently made. Tests of 
absorptive power for moisture are very often made. While 
the methods pursued by different people differ, they consist 
essentially of weighing the specimen dry, and then soaking it 
in either hot or cold water until it has absorbed all that it will, 
and weighing it again. 

There are a number of methods pursued in order to de- 



B U If. DINGS TONES. 7 1 5 



termine its power of withstanding the action of frost, and the 
results differ, of course, according to the method pursued. 
Bauschinger's method consists in — 

i°. Determining the compressive strength of the stone in 
a dry and in a saturated condition, and comparing the two. 

2°. Determining the compressive strength after twenty- 
five freezings and thawings. 

3 . Determining the loss of weight after these twenty-five 
freezings. 

4 . Examining the specimen with a microscope for cracks 
after the twenty-five freezings. 

Stones will not withstand the heat of a large conflagration, 
brick being better than any building-stone. 

As to how well a stone will stand the gases in the air of a 
large city, a great many tests have been proposed and used, but 
none of them are entirely satisfactory. Of course we can get 
indications from a study of the chemical composition, or better 
from a microscopical examination, which shows also the 
arrangement of the different components, this being, of course, 
the part of the geologist or mineralogist. 

After the question of durability, the strength comes in as 
the factor of next importance, and although the loads usually 
put upon stones in construction are very much smaller than 
the breaking-strength of the stone as shown by small speci- 
mens tested in the testing-machine, nevertheless the mortar 
or cement joints, the bonding, and the necessary unevenness 
render the real factor of safety very much less than would be 
at first imagined. 

Building-stones are more often called upon to bear a 
compressive load than any other, though they are sometimes 
called upon to bear a transverse load, as in the case of window- 
lintels. 

The results are very variable, partly because the stone 
varies very much in quality, and partly because it is only of 



?l6 APPLIED MECHANICS. 

late years that it has been recognized that in order to obtain 
correct results in compression tests the pressure must be evenly 
distributed over the surfaces of the specimen pressed upon r 
and that in order to accomplish this even distribution it is 
necessary that the faces which come in contact with the plat- 
forms of the testing-machine shall be accurate planes, and that 
unless the compression platforms are adjustable, the two faces 
pressed upon shall be parallel— provided, of course, the plat- 
forms are parallel, as they should be. Formerly it was thought 
that the desired result could be obtained by interposing be- 
tween the surface of the specimen and the platform some 
soft substance, as a cushion of wood or of lead, whereas it is a 
fact that any such cushions only render the results smaller 
and more variable than the real crushing-strength of the speci- 
mens. Hence it is that a great many of the tests that have 
been made are of no value, because this matter was not 
attended to. In a rough way we may divide the most com- 
mon building-stones as follows : i°. Granites and allied stones ; 
2°. Limestones, including marbles ; 3 . Sandstones ; 4 . Slates. 

The most systematic set of tests was made by Bauschinger, 
and is reported in the Mittheilungen, Hefte 1, 4, 5, 6, 10, 11, 
18, and 19. 

Besides this we may note the following references : 

i°. Report on Compressive Strength, etc., of the Building-Stones in 

the United States, 1876. Gillmore. 
2 . Compressive Resistance of Freestone, Brick Piers, Hydraulic 

Cements, Mortars, and Concretes, 1888. Gillmore. 
3 . Masonry Construction, 1889. Baker. 
4°. Testing Materials of Construction, 1888. Unwin. 
5 . History of the St. Louis Bridge, 1881. Woodward. 
6°. Exec. Doc. 12, 47th Congress, 1st session. Senate. 
7 . Exec. Doc. 35, 49th Congress, 1st session. Senate. 

Some tables of results will now be given. 



B UILD ING-S TONES. 



717 



The following is taken from Woodward's " History of the 
St. Louis Bridge :" 



Material. 



Grafton Magnesian Limestone. 



(5 specimens) 



Portland Granite 

<i << 

<« < 1 

Richmond Granite 

Portland Granite. 

Missouri Red Granite 

«c «« 

tc <( 

<< <« 

Brown Ochre Marble 

Sandstone, St. Genevieve, Mo 



Length, 

in 
inches. 



6.46 

5.87 
5-90 
5-99 
3.00 
8.00 
13.00 
5-88 



.98 

•97 

.00 

.00 

,00 

,00 

3.00 

3.00 

3.00 

3-oo 

4.88 

3.06 



Diameter of 

Cross-section, 

in inches. 



1. 14 
I.06 
I.06 
I.07 
3X3 
2.38 

1. 13 

2.36 
2.36 
2.38 
2.30 



X 3 
X 3 
X 3 
X 3 
X 3 
X 3 
X 3 
,88 X 4- 
06 X 3 



06 



Modulus of 

Elasticity, 

pounds per 

square inch. 



10500000 
8400000 
85000OO 
6000000 



I20000OO 
50000OO 
5 500000 
640OOOO 
5OO00O0 

I3500000 



Breaking- 
Weight, per 
square inch. 



7200 
8500 
2000 
6000 

av. 15400 

IOIOO 

10800 
16000 
18500 
17000 
16400 
13700 
12700 
13000 
12700 
13600 
15000 

5330 
5500 
3400 



In Heft 4 of the Mittheilungen, Bauschinger gives a long 
table of results of testing granites, limestones, and sandstones, 
from which the following examples are selected : 

In Heft 5 of the Mittheilungen is to be found a study of 
the modulus of elasticity of building-stones. Bauschinger 
found that in this case the departure from Hooke's law is 
greater with small than with large loads. Heft 6 contains an 
experimental study of the laws of compression. Heft 10 con- 
tains an experimental investigation of the principal Bavarian 
building-stones. Hefte 11 and 18 contain a study of the com- 
pressive strength and wearing qualities of paving-stones. Heft 
19 contains a study of the power of stones to resist frost. For 
all these the student is referred to the Mittheilungen. 



7 i8 



APPLIED MECHANICS. 



Kind of Stone. 



Place. 



Tests by Compres- 
sion. 



Crushing 

Strength, 

pounds 

per sq. in. 



Direction 
of the 

Pressure 
with re- 
spect to 
the bed. 



Transverse Tests, 



Modulus 
of Rup- 
ture, 
pounds 
per sq. in. 



Direction 
of Break- 
ing Sec- 
tion with 
respect to 
the bed. 



Granites. 
Yellow, very soft, exceptional quality 

Gray, coarse-grained 

Very light-colored, tolerably coarse- 
grained 

Very hard, coarse-grained .... 

Very hard, striped 

White, very hard, tolerably fine- 
grained, only good for paving . 

Syenite, black, with a good deal of 

gray 

Limestones. 

White marble 

Muschelkalk 

Yellowish white, soft limestone of the 
white Frankenjura 

Best quality 

Poorest quality 

Granitic marble 

Dolomite. 
From the white Frankenjura. . . . 



Sandstone. 

Bunterjsandstone, gray, with yellow 
and brown streaks 

Bunter sandstone, red, very rich in 

quartz, fine-grained 

Do. do. do. 

Bunter sandstone, dark red, fine- 
grained 

Do. do. do. 

Keuper sandstone, red, fine-grained. 

Do. do. do. 

Keuper sandstone, white, with red- 
dish bed stripes, coarse-grained. 

Keuper sandstone, white, tolerably 
fine-drained 

Keuper sandstone, brown .... 

Green sandstone, yellow, with brown 
layers, fine-grained 

Green sandstone, dirty green, tolera- 
bly fine-grained 

Green sandstone, greenish, fine- 
grained 



Selb in Ober- 
f ran ken. 



Waldstein 
St. Gotthard 



Cham 

Ebendaher 

Schlanders in 

Tyrol 

Wurzburg 

Kronach 

Kelheim 



Rosenheim 

Poppenheim 
Herzbruch 

Kronach 
Unterfranken 

Carlsruhe 
Wurtemburg 

Ansbach 
(i 

Nuremberg 

Regensberg 
(( 

Kelheim 



775° 
1 1300 
1 1720 

14790 
1 1 740 
12660 
15650 
13230 

22190 

19200 



12800 

6260 

22760 

1 1390 



6900 

8390 

3560 

10600 

20340 

17920 

18490 
16780 
12520 
1 1 240 



4836 

19270 
20550 

1 1950 
8100 



939o 
654c 



2490 
434o 

5480 

2680 
4410 

3630 



av.1940 
1309 
2773 



1991 



oblique 

II 
across 



1252 
939 



2560 



B UILDING-S TONES. 



719 



The following table is taken from the Trans. Am. Soc. Civ. 
Eng. for Oct. 1886, where it is quoted from " Mechanical tests 
of building materials, made at the Watertown Arsenal, by the 
United States Ordnance Dept., at the request of the Commis- 
sioners for the erection of the Philadelphia Public Buildings :" 









°<u 


Total 

Load 

applied, 

lbs. 


Crushing 






C 

.BCn 


Locality. 


Color. 


g 

'« 03 
U V 

<U U 

Q 


Strength 

per sq. in. 

applied, 

lbs. 


"3 

.2 « . 

CO 


Remarks. 


r 


Lee, Mass 


Blue. 


End. 


715000 


20504 


34.87 


Burst in fragments. 




" " .... 


White. 


Bed. 


800000 


22370 


35- 16 


Slight flaking. 




44 " .... 


W. &B. 


End. 


800000 


22860 


34-99 


No apparent injury. 




" " .... 


White. 


" 


800000 


22820 


35-05 


" " " 




" " .... 


Blue. 


Bed. 


800000 


22900 


34-93 


Flaked, one edge. 


X! 


44 :4 .... 


W. &B. 


" 


767000 


21700 


35-34 


Crushed suddenly. 


15 


Montgomery Co., Pa. 


Blue. 


" 


466300 


1 1470 


40 64 


Failed suddenly. 


S 


n a 


" 


End. 


400000 


10420 


38.40 


Ultimate strength. 




It II 


" 


Bed. 


543000 


13700 


39-63 


it n 




„ 


ii 


End. 


398000 
347500 


10120 
959o 


39-33 
36.24 


.. 


I 


" " 


" 


Bed. 


434000 


10940 


39.67 


" " 


A 


Conshohocken, Pa. . 




End. 


494000 


14090 


35-05 


Ultimate strength. 


si it 




Bed. 


566000 
377000 


16340 
8530 


34-63 
44.22 


a it 


II 


Indiana 




End. 


it <i 


II 

. 1 

V 

2 


11 




" 


320500 


7190 


44-56 


1. »i 


11 




Bed. 


321000 


7776 


41.38 


" " 


» 




" 


438300 
531200 




41.28 
39 65 


ii ti 


Vermont J 




Dove- | 
colored > 


Bed. 


13400 


Ultimate strength. 


End. 


379800 


9870 


38.48 


it it 



•u o 



Locality. 



Hummelstown, Pa. 
Ohio 





°<; 




Crushing 






3 




Strength 


1 - c ' 


Color. 


u S 

V v. 


applied, 
lbs 


per sq. in. 
applied, 


rt-H 

.2 « • 




p 




lbs. 


in 




Bed. 


528700 


12810 


41.28 




End 


570300 
256000 


13610 
6510 


41.92 
39.32 


Buff. 


Bed. 


" 


End. 


199500 


4860 


41.02 


" 


Bed. 


289500 


7020 


41-25 


tt 


End. 


160000 


394o 


40.06 


Blue. 


44 


305000 


7680 


39-68 


" 


Bed. 


435400 


10400 


41.90 


" 


End. 


391800 


9795 


40.00 


" 


Bed. 


351000 


8710 


40.30 


" 


End. 


672 TOO 


16280 


41.28 




Bed. 


493500 


17420 


39-74 



Remarks. 



Ultimate strength. 

ti 11 

it ti 

11 <i 



Bearings imperfect. 
Ultimate strength. 



Fractured suddenly. 
Ultimate strength. 



7 2 ° APPLIED MECHANICS. 

§ 243. Hydraulic Cements and Brick Piers.— When a 

pure or nearly pure limestone is calcined, so as to drive off the 
carbonic acid, we have an oxide of lime, commonly called quick- 
lime, which, on the addition of water, slakes, with the develop- 
ment of considerable heat; and the result is a fine powder, which 
by the addition of more water is reduced to a paste which slowly 
hardens upon exposure to the air. It is this paste, mixed with 
sand, which forms the mortar used in cheap buildings. It is very 
weak, and hardens very slowly, even in the air. 

On the other hand, a hydraulic lime or a hydraulic cement 
contains impurities, of which silica forms the principal portion, 
though we usually find also alumina, protoxide of iron, and 
magnesia; and these impurities are in so large a proportion that 
the slaking entirely or nearly disappears, and the addition of 
water after calcination causes the formation of hydrated silicates, 
etc., which harden under water. 

While we cannot draw a sharp line of demarcation between 
hydraulic lime and hydraulic cement, nevertheless the essential 
difference is that the first contains pure lime in sufficient pro- 
portions to slake, but at the same time contains enough clay, 
silica, etc., to enab>le it to set under water; whereas hydraulic 
cement contains less pure lime, and hardly slakes at all, but 
sets more rapidly than hydraulic lime. 

Hydraulic cements are known as, i°, Portland cement, and, 
2 , Natural cement. The latter is commonly called Rosendale 
in America, and Roman in Europe. It acquires its strength 
more slowly, is weaker and cheaper than Portland, and it usually 
sets more quickly. 

Portland cement is manufactured extensively in France, Ger- 
many, and England, and in the United States. It is made by 
mixing either dry or in paste, and then calcining, to the point of 
incipient vitrefaction, such mixtures of rocks as will give the 
proper chemical composition. The paste, when ready for the 
furnace, should contain from 76 to 81 per cent of carbonate of 
lime, and from 19 to 24 per cent of clay. 



HYDRAULIC CEMENTS AND BRICK PIERS. 721 

To give precise definitions of what constitutes Portland 
cement, what Natural cement, what Hydraulic lime, etc., is not 
an easy matter. An attempt to do so was made by the Inter- 
national Association for Testing Materials, but their definitions 
are not universally accepted, and will not be given here. 

For definitions of Portland cement, and of Natural cement, 
which are by no means perfect, but which will answer in a general 
way, the reader is referred to those adopted by the Am. Soc. for Test- 
ing Materials on pp. 730 and 733. The manufacture of Natural 
cement dates from a very early period, and depends upon finding 
rocks of suitable composition; that of Portland cement dates 
from the early part of the nineteenth century, when Jos. Aspdin, 
of Leeds, made a slow-setting cement, by calcining a mixture of 
carbonate of lime, and clay, in suitable proportions. 

TESTS OF THE STRENGTH OF CEMENTS. 

While a good many tests have been made on the compres- 
sive strength of cement, and a few also on transverse or shear- 
ing strength, nevertheless the test most commonly used in 
order to determine its quality is the test of its tensile strength. 
The specimen used for this purpose is called a briquette, and 
the cut shows one of its common forms, the smallest 
section being generally one square inch. The real 
reason for using the tensile instead of the compres- 
sive strength for a test is, in the opinion of the 
author, that inasmuch as the tensile strength of 
cement is very much less than its compressive 
strength, it follows that the machines for testing the 
tensile strength are cheaper, and the work of testing tensile 
strength is less than is the case in testing its compressive 
strength, although there are some who give other reasons 
which have some appearance of plausibility. 

In order to discuss this matter intelligently, however, we 
should bear in mind — 

Either the tensile or the compressive test is compara- 




72 2 APPLIED MECHANICS. 



tive and merely helps to determine the quality, and also to 
compare different lots of cement, but neither of them furnish 
us any figures which would be suitable to use in computing 
the allowable load on any structure which depended upon 
cement for its strength. 

We might, therefore, conclude that, as far as the objects 
that can be attained by testing cement are concerned, either 
test would answer the purpose equally well; but inasmuch 
as it is also a fact that, with the best appliances thus far 
provided for the purposes, it is possible to obtain greater 
accuracy in the compressive than in the tensile test, therefore 
it seems to the author that the compressive and not the tensile 
is the test that should be used in making cement tests. Never- 
theless, inasmuch as the tensile strength is most used, a brief 
account will be given here, showing what has been done, what 
we can reasonably expect from good cements, and a few pre- 
cautions will be mentioned, which it is necessary to use in 
making the tests, in order to insure correct results. 

The literature of cement testing is very extensive, but only 
ihe following will be given here 

i°. Q. A. Gillmore: Practical Treatise on Limes, Hydraulic Cements, 
and Mortars. 

2°. John Grant: Articles in the Proceedings of the British Institu- 
tion of Civil Engineers, vols, xxv., xxxii., and xli. 

3°. Charles Colson: Experiments on the Portland Cement used in 
the Portsmouth Dockyard Extension. Proc. Brit. Inst. Civ. 
Engrs., vol. xli. 

4°. Isaac John Mann: The Testing of Portland Cement. Proc. Brit. 
Inst. Civ. Engrs., vol. xlvii. 

5°. Wm. V. Maclay: Notes and Experiments on the Use and Testing 
of Portland Cement. Trans. Am. Soc. Civ. Engrs., Dec. 1877. 

6° Eliot C. Clarke: Record of Tests of Cement made for the Bos- 
ton Main Drainage Works, 1878-1884. Trans. Am. Soc. Civ. 
Engrs., April, 1885. 



HYDRAULIC CEMENTS AND BRICK PIERS. 72$ 

7°. Q. A. Gillmore: Notes on the Compressive Resistance of Free- 
stone, Brick Piers, Hydraulic Cements, Mortars, and Con- 
cretes. 

8°. J. Sondericker: How to Test the Strength of Cements. Am. 
Soc. Mech. Engrs. for 1888. 

9°. Bauschinger: Mittheilungen aus dem Mechanisch-Technischen 
Laboratorium, Hefte i., vii., and viii. 

io°. Exec. Doc. 12, 47th Congress, 1st session, House: Compres- 
sive Tests of Seven Cubes of Concrete. 

ii°. Exec. Doc. 5, 48th Congress, 1st session, Senate: Shearing 
Test of One Concrete Cube. 

12 . Exec. Doc. 35, 49th Congress, 1st session, Senate: Tests of 
Neat Cement and Cement Mortars. 

13°. Preliminary Report of the Committee on a Uniform System for 
Tests of Cement. Trans. Am. Soc. Civ. Engrs., January, 1884. 

14 . Final Report of the Committee on a Uniform System for Tests 
of Cement. Trans. Am. Soc. Civ. Engrs., January, 1885. 

1 5 . Behavior of Cement Mortars under various Contingencies of 
Use. F. Collingwood: Trans. Am. Soc. Civ. Engrs., Nov. 
1885. 

1 6°. Report of Progress by the Committee on the Compressive 
Strength of Cements, etc. Trans. Am. Soc. Civ. Engrs., July, 
1886. 

1 7 . Another Report of the Committee. Trans. Am. Soc. Civ. 
Engrs., June, 1888. 

18 . E. F. Miller : Testing Cement. The Bricklayer. 

1 9 . Candlot, E. : Ciments et Chaux hydrauliques. Fabrication, 
Proprietes, Emploi. Paris. 

20 . Feret, R. : Note sur Diverses Experiences concernant les 
Ciments. Annales des Ponts et Chaussees, 1890, i r semestre, 
page 313. 

21 . Alexandre, Paul : Recherches experimentales sur les Mortiers 
hydrauliques. Annales des Ponts et Chaussees, 1890, 2 e se- 
mestre p. 227. 

22 . Feret, R. : Sur la compacite de Mortier hydraulique. Annales 
des Ponts et Chaussees, 1892, 2 e semestre, page 5. 



724 APPLIED MECHANICS. 

23 . D. B. Butler: Portland Cement. 

24 . Baumaterialienkunde : This is the official organ of the Inter- 
national Assoc, for Testing Materials, and contains many 
papers, and discussions on cement. 

2 5 . Commission des methods d'essai des materiaux de construc- 
tion. Tome 1, Section B. — Essais des materiaux d'aggre- 
gation des maconneries. Rapport General presente par 
Paul Alexandre. 

2 6°. Tests of Metals made at Watertown Arsenal. 

2 7 . Mit . der Materialprufungsanstalt in Zurich. 

28 . Mitt, aus dem Mech. Tech. Lab. in Berlin. 

29 . Mitt, aus dem Mech. Tech. Lab. in Miinchen. 

30 . Report of Board of Engineer officers on testing hydraulic 
cements, 1902. 

31 . Many articles in the Trans. Am. Soc. Civil Engineers. 

3 2 . Many papers read before the Association of American Portland 
Cement Manufacturers. 

Some quotations will be given from Candlot's treatise, 
including a portion of the specifications of the French Maritime 
Service. 

Candlot says : 

" The properties of Portland cement, which have given complete 
satisfaction for many years, being known, well denned, and absolutely 
constant, it ought to be sufficient, in order to determine the value of a 
cement, to see whether it presents, to the same degree, the qualities 
which characterize this list of hydraulic products." 

" The tests generally made on cements have to do with their chemical 
composition, their density, their fineness of grinding, their time of set- 
ting, their tensile and compressive strength, and their invariability of 
volume." 

The reason for each of these tests is so plain that no comment will 
be made, except to say that a cement that swells is liable to disintegrate 
after setting in consequence of free lime. 

In France the greater part of the large manufactories are to be found 
in the region around Boulogne-sur-Mer ; and the maritime service of 
the Department of " Ponts et Chauss6es," which has a cement laboratory 
at Boulogne, has established certain specifications to which all cement 
used in their work must conform. 



HYDRAULIC CEMENTS AND BRICK PIERS. 72$ 

A portion of these specifications as given by Candlot will now be 
quoted in the following three pages : 

Art. i. The Portland cement furnished shall come exclusively from 
the manufactory of the one who offers it for sale. It shall be produced 
by grinding scorified rocks, obtained by calcining to the point of vitrifi- 
cation, of an intimate mixture of carbonate of lime and clay, carefully 
mixed, and chemically and physically homogeneous throughout. 

Art. 2. The administration reserves the right, under conditions 
which it determines, to supervise the manufacture, the storing at the 
factory, and the shipping of the cement. 

For this purpose the engineer or his representative shall have access 
at all times to all parts of the factory concerned ; and he may — 

i°. Do whatever he thinks necessary to make sure of the composi- 
tion of the crude pastes used. 

2°. Supervise the sorting after calcining. 

3°. Follow the cement after the sorting to the special cases where it 
is to be stored after grinding. 

4°. Supervise the packing when it is taken from the cases, and also 
the shipping of the cement. 

5°. Place special agents permanently at the factory for the above- 
stated purposes. 

Art. 4. Every partial lot of cement, on its arrival at the storehouse 
of the works, must be examined as to dryness. No bag shall be allowed 
to enter which has been exposed to dampness, or whose contents is not 
entirely pulverulent throughout. Then the part allowed to enter, as far 
as dryness is concerned, shall be submitted to the tests prescribed for, 
1°, density; 2°, chemical composition ; 3 , time of setting; 4 , absence 
of cracks after setting; 5 , strength of briquettes of neat cement; 6°, 
strength of briquettes of cement with normal sand. 

The engineer, or his representative, shall take some cement from one 
or more bags chosen arbitrarily at such points as he shall decide, but 
without mixing cement from different bags. He shall then proceed to 
the tests, observing the precautions prescribed. Each of the specimens 
thus chosen must satisfy separately the conditions prescribed ; the 
measures to be taken in regard to the whole of a partial lot being those 
suitable for the specimen giving the least satisfactory result. 

Art. 5 gives very elaborate instructions in regard to the determination 
of the minimum weight per litre of cement that has passed a sieve of 5000 
meshes per square centimetre. A portion of this article is as follows, 
viz.: — To obtain, under conditions always comparable, an unheaped litre 



7 26 APPLIED MECHANICS. 

of the fine dust, produced by sifting cement through a sieve of 5000 
meshes per square centimetre, we place on a firm support a measure of 
one litre capacity ; above this measure we arrange a plane inclined at 
45° formed of a sheet of zinc 50 cm. long, whose horizontal lower edge 
shall be fixed one centimetre above the level of the upper plane of the 
measure ; we pour, gently, the cement dust, by means of a spoon, onto 
the inclined plane at the top, until the measure is a little more than 
filled, and we remove the excess of cement by sliding over the edges of 
the measure (must not be subjected to jar or shock). To obtain the 
weight of a litre, we make one single weighing of the total amount of 
five measures, filled with the above-described precautions. 

Art. 6. Every cement in which the chemical analysis shall show 
more than 1% sulphuric acid or compounds of sulphur in measurable 
proportion shall be rejected. 

Art. 7. All cement will be declared suspected in which chemical 
analysis shows more than 4% of oxide of iron, or which has a value less 
than -^ for the ratio between the total weight of the combined silicon 
and aluminum, on the one hand, and the lime on the other, 
more than 1% sulphuric acid or compounds of sulphur in measurable 
proportion shall be rejected. 

Art. 8. In the tests of neat cement, the cement shall be mixed in 
sea-water. The water and air during mixing shall be kept as nearly as. 
possible between 15 and 18 C. 

To determine the proper proportion of water to mix with the cement 
we make the following preliminary test : 

The mortar is obtained by taking 900 grammes of cement, and pour- 
ing on it all at once the water to be used, mixing the mortar with a 
trowel on a marble slab for five minutes from the moment of pouring; 
the water. 

The quantity of water used shall be considered normal if the mortar 
forms a firm paste, well united, brilliant and plastic, satisfying the follow- 
ing conditions: 

i°. The consistency of the paste must not change if the mixing goes 
on for eight minutes instead of five. 

2 . A small quantity of paste taken with the trowel and let fall on the 
marble from about 50 cm. must detach itself from the trowel without 
leaving any adhering to the trowel, and, after its fall, it must preserve 
approximately its form without cracking. 

3 . A small quantity of paste being taken in the hand, it must be 
sufficient to give it some light taps to give it a rounded form and to- 
make the water come to the surface ; it must neither flatten out com- 



HYDRAULIC CEMENTS AND BRICK PIERS. J2J 

pletely nor stick to the skin, and if the ball be let fall from half a metre, it 
must preserve a rounded form (slightly flattened), without cracks. 

4°. With less water the paste should be dry, not well united, and 
should show cracks in falling. With more water it should have a muddy 
consistency, with adherence to the trowel. 

After making a series of successive approximations we must adopt as 
normal proportion the greatest proportion of water tried which shall 
have produced a plastic and not a muddy paste satisfying the conditions 
stated. 

Art. 9. A part of this article reads as follows : With a part of the 
paste thus obtained we fill a cylindrical metal box of 0.04 m. height and 
0.08 m. diameter, jarring it a few seconds, and leaving the water that 
rises to the top. Then suspend, by a cord passing over a pulley, a Vicat 
needle of 300 grams weight and a square section 1 mm. on a side, and 
lower it gradually. 

The beginning of the set is taken as the time when the needle ceases 
to penetrate to the bottom of the mould, and the end of the set, as the 
time when the needle no longer penetrates appreciably. 

Times are estimated from the moment when the water is poured on 
the dry powder. 

If the cement begins to set before thirty minutes or completes its set 
before three hours, the partial lot shall be rejected ; the temperature 
during the operation having been between 15 and 18 C. 

Art. 10 prescribes a form of test to guard against the presence of 
cracks after setting. 

Art. 11. The paste for tensile tests of neat cement is obtained by 
mixing with a trowel, on a marble slab, during 5 minutes, 900 grammes 
of cement with the normal quantity of water, as already determined. 
Each mixing will furnish paste for 6 briquettes. Make three successive 
mixings to obtain 18 briquettes, which is the number to be used in each 
test. 

The form of the briquette is prescribed, the thickness being o m .o222, 
the smallest section being o m .o225 wide; area, 5 square centimetres. 

Put the moulds on a marble slab, and fill each set of six with one 
mixing, putting enough in each mould at once so that it shall overflow. 
Pack with the flat of the trowel. When the filling is complete, give 
little taps with the trowel handle on the side to disengage bubbles 
of air. 

As soon as the consistency of the cement permits, smooth off the 
upper surface even with the mould by using the blade of a knife. 



728 APPLIED MECHANICS. 

After the cement has set remove the moulds, leaving the briquettes 

on the slab. 

During the first 24 hours the briquettes must be kept on the slab, in 
a damp atmosphere, free from currents of air and the direct rays of the 
sun, at a temperature of from 15 to 18 C. 

After 24 hours immerse them in sea-water, the water to be renewed 
every week, and kept, as nearly as possible, at a temperature between 
15 and 1 8° C. 

For each sample of cement to be tested make 18 briquettes of neat 
cement, of which 6 are to be broken 7 days from the time of mixing, 6 
at the end of 28 days, and 6 at the end of 84 days. For each series take 
one briquette from each mixing. 

The testing-machine prescribed is one where the tension is obtained 
by pouring a jet of grains of lead into a vase at the end of a second 
lever. Among the six results in each series, choose the three highest; 
the mean of these three shall be considered to be the strength of the 
sample tested at that time. 

Art. 12. h. The resistance of briquettes of neat cement at the end of 
the 7th day must be at least 20 kilogrammes per square centimetre. It 
must be at least 35 kg. at the end of the 28th day. Every partial lot 
whence comes a sample not satisfying these two conditions shall be 
rejected. 

Art. 13. The strength per square cm. at the end of 28 days must be 
at least 5 kg. greater than that at the end of 7 days ; otherwise the 
partial lot shall be suspected, the suspicion not to be removed unless the 
strength at the end of 28 days is at least 55 kg. 

Art. 14. The strength per square cm. at the end of 84 days must be 
at least 4$ kg. It must also exceed the strength at the end of 28 days 
when the latter was not at least 55 kg. Every partial lot not satisfying 
these conditions to be rejected. 

The tests of cement mortar are made on briquettes of mortar com- 
posed of one part by weight of cement to three of normal sand, the 
latter being furnished by the Administration, and being such as will pass 
through a sieve of 64 meshes per square centimetre and be rejected by 
one of 144 meshes per square centimetre. 

The amount of water used is 12% of the total weight of cement and 
sand. 

Very minute directions are given in regard to the mixing and pre- 
paring the briquettes very similar to those for neat cement, and then 
the specifications proceed as follows, viz.: For each sample of cement 



HYDRAULIC CEMENTS AND BRICK PIERS. 729 

we make 18 briquettes of normal sand mortar, of which 6 are to be 
broken at the end of 7 days, 6 at the end of 28 days, and 6 at the end of 
84 days ; using in each series a briquette from each of the six different 
mixtures in which the mortar is to be made. Of the six results in each 
series we take the three highest, and the mean of these is the figure ad- 
mitted iOr Lhe resistance ^f ..ne mortar. 

Art. 17. c. The strength of normal sand mortar at the end of seven 
days must be at least 8 kg. per sq. cm., and at the end of 28 days at 
least 15 kg. per sq. cm. Each partial lot whence comes a sample not 
satisfying these conditions is to be rejected. 

Art. 18. The resistance at the end of 28 days must exceed that at 
the end of 7 days by at least 2 kilogrammes, otherwise the partial lot is 
to be suspected. 

Art. 19. The resistance at the end of 84 days must be at least 18 
kilogrammes, and it must exceed the resistance at the end of 28 days. 
Every partial lot whence comes a sample not satisfying these conditions 
should be rejected. 

The German, the Swiss, and other specifications may be 
found in Candlot's book; but a portion of those of the American 
Society for Testing Materials will be quoted here. 

AMERICAN SOCIETY FOR TESTING MATERIALS. 

REPORT OF COMMITTEE ON STANDARD SPECIFICATIONS FOR 

CEMENT. 

General Observations. 

1. These remarks have been prepared with a view of pointing out 
the pertinent features of the various requirements and the precautions 
to be observed in the interpretation of the results of the tests. 

2. The Committee would suggest that the acceptance or rejection 
under these specifications be based on tests made by an experienced 
person having the proper means for making the tests. 

3. Specific Gravity. — Specific gravity is useful in detecting adultera- 
tion or underburning. The results of tests of specific gravity are not 
necessarily conclusive as an indication of the quality of a cement, but 
when in combination with the results of other tests may afford valuable 
indications. 

4. Fineness. — The sieves should be kept thoroughly dry. 

5. Time of Setting. — Great care should be exercised to maintain 
the test pieces under as uniform conditions as possible. A sudden 



73° APPLIED MECHANICS, 

change or wide range of temperature in the room in which the tests are 
made, a very dry or humid atmosphere, and other irregularities vitally 
affect the rate of setting. 

6. Tensile Strength. — Each consumer must fix the minimum re- 
quirements for tensile strength to suit his own conditions. They shall, 
however, be within the limits stated. 

7. Constancy of Volume. — The tests for constancy of volume are 
divided into two classes, the first normal, the second accelerated. The 
latter should be regarded as a precautionary test only, and not infallible. 
So many conditions enter into the making and interpreting of it that 
it should be used with extreme care. 

8. In making the pats the greatest care should be exercised to avoid 
initial strains due to molding or to too rapid drying-out during the first 
twenty-four hours. The pats should be preserved under the most 
uniform conditions possible, and rapid changes of temperature should 
be avoided. 

9. The failure to meet the requirements of the accelerated tests 
need not be sufficient cause for rejection. The cement may, however, 
be held for twenty-eight days, and a retest made at the end of that 
period. Failure to meet the requirements at this time should be con- 
sidered sufficient cause for rejection, although in the present state of our 
knowledge it cannot be said that such failure necessarily indicates 
unsoundness, nor can the cement be considered entirely satisfactory 
simply because it passes the test. 

General Conditions. 
Of these the first eight will not be quoted here. 

9. All tests made in accordance with the methods proposed by 
the Committee on Uniform Tests of Cement of the American Society 
of Civil Engineers, presented to the Society, January 21, 1903, and 
amended January 20, 1904, with all subsequent amendments thereto. 
(See addendum to these specifications.) 

10. The acceptance of rejection shall be based . on the following 
requirements: 

Natural Cement. 

11. Definition. — This term shall be applied to the finely pulverized 
product resulting from the calcination of an argillaceous limestone at a 
temperature only sufficient to drive off the carbonic acid gas. 



AMERICAN SOCIETY FOR TESTING MATERIALS. 731 

12. Specific Gravity. — The specific gravity of the cement thoroughly 
dried at ioo° C, shall be not less than 2.8. 

13. Fineness. — It shall leave by weight a residue of not more than 
10 per cent on the No. 100, and 30 per cent on the No. 200 sieve. 

14. Time of Setting. — It shall develop initial set in not less than ten 
minutes, and hard set in not less than thirty minutes, nor more than 
three hours. 

15. Tensile Strength. — The minimum requirements for tensile 
strength for briquettes one inch square in cross-section shall be within 
the following limits, and shall show no retrogression in strength within 
the periods specified:* 

Age. Neat Cement. Strength. 

24 hours in moist air 50-100 lbs. 

7 days (1 day in moist air, 6 days in water) 100-200 " 

28 " (I " " " " 27 " " u ) 200-300 " 

One Part Cement, Three Parts Standard Sand. 

7 days (I day in moist air, 6 days in water) 25- 75 lbs. 

28 " (1 " " " " 27 » « » ) 75- I 5° " 

16. Constancy of Volume. — Pats of neat cement of about three 
inches in diameter, one-half inch thick at center, tapering to a thin 
edge shall be kept in moist air for a period of twenty-four hours. 

(a) A pat is then kept in air in normal temperature. 

(b) Another is kept in water maintained as near 70 F. as prac- 
ticable. 

17. These pats are observed at intervals for at least 28 days, and, 
to satisfactorily pass the tests, should remain firm and hard and show 
no signs of distortion, checking, cracking or disintegrating. 

Portland Cement. 

18. Definition. — This term is applied to the finely pulverized 
product resulting from the calcination to incipient fusion of an intimate 
mixture of properly proportioned argillaceous and calcareous materials, 
and to which an addition no greater than 3 per cent has been made 
subsequent to calcination. 

* For example the minimum requirements for the 24-hour neat cement test 
should be some value within the limits of 50 -and 100 lbs., and so on for each 
period stated. 



73 2 APPLIED MECHANICS. 

19. Specific Gravity. — The specific gravity of the cement, thor- 
oughly dried at ioo° C, shall not be less than 3.10. 

20. Fineness. — It shall leave by weight a residue of not more than 
8 per cent on the No. 100, and not more than 25 per cent on the No. 
200 sieve. 

21. Time of Setting. — It shall develop initial set in not less than 
thirty minutes, but must develop hard set in not less than one hour, nor 
more than ten hours. 

22. Tensile Strength. — The minimum requirements for tensile 
strength for briquettes one inch square in section shall be within the 
following limits, and shall show no retrogression in strength within the 
periods specified : * 

Age. Neat Cement. Strength. 

24 hours in moist air 150-200 lbs. 

7 days (1 day in moist air, 6 days in water) 450-550 " 

28 " (1 " " " " 27 " " " ) 550-650 " 

One Part Cement, Three Parts Standard Sand. 

7 days (1 day in moist air, 6 days in water) 150-200 lbs. 

28 " (1 " " " 4< 27 " " " ) 200-300 " 

• 

23. Constancy oj Volume. — Pats of neat cement about three inches 
in diameter, one-half inch thick at the center, and tapering to a thin 
edge, shall be kept in moist air for a period of twenty-four hours. 

(a) A pat is then kept in air in normal temperature and observed 
at intervals for at least twenty-eight days. 

(b) Another pat is kept in water maintained as near 70 F. as 
practicable* and observed at intervals for at least twenty-eight days. 

(c) A third pat is exposed in any convenient way in an atmosphere 
of steam, above boiling water, in a loosely closed vessel for five hours. 

24. These pats, to satisfactorily pass the requirements, shall remain 
firm and hard and show no signs of distortion, checking, cracking or 
disintegrating. 

* For example the minimum requirement for 24-hour neat cement test should 
be some value within the limits of 150 and 200 lbs., and soon for each period 
stated. 



HYDRAULIC CEMENTS AND BRICK PIERS. 733 



25. Sulphuric Acid and Magnesia. — The cement shall not contain 
more than 1.75 per cent of anhydrous sulphuric acid (SO3), nor more 
than 4 per cent of magnesia (MgO). 

Bauschinger has also made a large number of compression 
tests, accounts of which may be found in Hefte 1, 7, and 8 of the 
Mittheilungen, but for these the student is referred to the Mittheil- 



PRECAUTIONS TO BE OBSERVED IN TESTING CEMENTS. 

The results obtained by testing different samples of the same 
cement will vary with — 

i°. The percentage of water used in mixing. 

2 . The length of time the sample has been kept under water 
and also the length of time it has been kept in the air before 
testing. 

3 . The temperature of the water with which it was mixed, 
and also of that in which it was kept; also the temperature of the 
air in which it was kept. 

4 . The rapidity of breaking. 

Hence, in order that our results may be of value, we must 
take pains to regulate all these matters. 

But another and all-important matter that has not received the 
necessary amount of attention is, that some means should be 
adopted for distributing the pull, in the case of a tension test, 
evenly over the section of the briquette, and in the case of a com- 
pression test for distributing the thrust evenly over the surface 
of the specimen. In the ordinary cement-testing machines to 
be found in the market there is generally no adequate provision 
for this purpose, and this is the reason why so great a variation 
exists in the results obtained with the same cement by so many 
experimenters. For a fuller account of this matter see Trans. 
Am. Soc. Mech. Engrs. for 1888, page 172. 

The following is a summary of a part of a paper read by Mr. 



?34 APPLIED MECHANICS. 

James E. Howard of Watertown Arsenal, before the Assoc, of 
Am. Portland Cement Mfr., in April, 1905. He says: 

i°. That a 100-mesh sieve has openings o."oo58 diam. 

That a 200-mesh sieve has openings o/'oo3i diam. 

That a No. 20 bolting cloth has openings o '/■ '0027 diam. 

He advises the use of the latter for the separation of the fine 
from the coarse particles. 

2 . That while he obtained for freshly ground Portlands 
specific gravities in the vicinity of 3.1, there were a number of 
natural cements examined, which had substantially the same 
values as Portlands, although some brands fell below 3. 

That hydration, partial or complete, lowers the specific gravity. 
That hydration begins at once, goes on more quickly in the finer 
particles and more slowly in the coarser ones. That, in some 
cases, hydration was not complete at the end of five days. Hence, 
that the usual arbitrary methods of determining the beginning 
and the end of the set do not show the beginning and end of 
hydration. 

That, as a rule, the compressive strength of the cement will not 
be diminished, if a period of about eight hours intervene between 
gauging and use. 

3 . That exposure to high temperatures is liable to lead to 
ultimate disintegration. 

4 . That a number of compression specimens were moulded 
under pressures varying from 7000 to 14000 pounds per square 
inch, continued for 40 hours or more, and were subsequently 
tested, at ages of 1 and 2 months. They developed phenomenal 
strength, a sample of neat cement showing a strength of 22050 
pounds per square inch at the age of 57 days, and 1 to 1 mortar 
191 20 pounds per square inch at the age of 1 month. 

Setting under high pressures admits of the use of smaller 
quantities of water in gauging; in one case only 5 per cent having 
been used. 



HYDRAULIC CEMENTS AND BRICK PIERS. 7$$ 



TESTS OF FULL-SIZE PIECES. 

Inasmuch as cement and mortar are almost always used as 
binding materials, tests of full-size pieces in which they enter are 
those of some form of masonry. Of such tests the number is not 
large, and those that will be quoted here are some tests of reen- 
forced concrete, and some of brick pieces. 

Concrete is composed of mortar and some hard material, as 
gravel, broken stone, cinder, etc., the general plan being to so 
proportion them that the cement shall approximately fill the voids 
in the sand, and that the mortar shall approximately fill the voids 
in the broken stone, or other hard material used. 

Reenforced concrete, which is made by imbedding in the 
concrete, iron or steel bars, wire mesh or expanded metal, etc., 
is now attracting a great deal of attention. 

COLUMNS. 

An extensive series of tests of columns of reenforced concrete 
is now being carried on at the Watertown Arsenal, and the follow- 
ing table, which gives a summary of the tests of this kind already 
published, is quoted from "Tests of Metals for 1904." 



73<5 



APPLIED MECHANICS. 



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HYDRA ULIC CEMENTS AND BRICK PIERS. 



717 



The following table gives the results of a set of tests made in 
the Laboratory of Applied Mechanics of the Mass. Institute of 
Technology, upon reenforced concrete columns, the concrete 
consisting of i part Portland cement (Star brand), 2 parts sand, 
and 6 parts trap-rock. 



V* 

V 

JD 

5 


cS 


P 

Gc/2 


biiju 


'0 


6 


s 

'•Sri 

.§•8 


• - <u 

*c3H 


M 

<A O 


Remarks. 




c 


tl 


fc 


03 


S 


m 








In. In. 














I 


30 


8X8 


17 


1 


I 


P 


107000 


Crushed at end. 


2 


30 


' ' 


17 


1 


I 


T 


127000 


Buckled, then crushed at end. 


3 


29 


" 


12 


1 


I 


P 


1 00000 


Buckled, then crushed at end. 


4 


28 




12 


1 


I 


T 


126000 


Crushed at end. Poorly made. Crushed 
portion cut off; the rest bore 150000 
lbs. at 40 days. 


5 


32 




6 


1 


I 


P 


138000 


Crushed at middle, then sheared off 
along rod to end. 


6 


31 


' ' 


6 


1 


I 


T 


133000 


Crushed at end. 


7 


31 


" 


17 


1 


li 


P 


136000 


Crushed at end, shearing obliquely. 


8 


25 


' ' 


17 


1 


I* 


T 


154000 


Crushed at end, breaking off 3 feet. 


9 


35 


■ ' 


17 


4 


i 


P 


182000 


Crushed at end. 


10 


34 




17 


4 


i 


T 


167000 


Crushed at end, concrete rather poor 
and rough at that end. 


11 


31 


' ' 


12 


4 


* 


T 


147000 


Crushed and split open at end. 


12 


32 


' ' 


12 


4 


i 


P 


153000 


Crushed and split open at end. 


13 


29 


1 ' 


6 


4 


1 


T 


158000 


Crushed and split open at end. 


14 


31 


■ ' 


6 


4 


1 


P 


244000 


Crushed at end. 


IS 


35 


10 X 10 


17 


1 


1 


P 


215000 


Broke off clean for 3 or 4 feet at end. 


16 


35 


' ' 


6 


1 


1 


P 


240000 


Sheared diagonally at end. 


17 


45 




6 


1 


1 


T 


228400 


Sheared diagonally at end, and broke 
back for half the length. 


18 


3i 


' ' 


12 


1 


1} 


T 


262000 


Sheared diagonally near end. 


19 


29 


' ' 


12 


1 


i* 


P 


257000 


Crushed at end. 


20 


28 


* ' 


12 


4 


f 


T 


300000 


Not broken. 


21 


29 




12 


4 


i 


P 


274000 


Crushed at end. Wedge-shaped piece 
forced in between rods. 



REENFORCED CONCRETE BEAMS. 

While more or less theorizing has been done by different 
people regarding suitable formulae for use in the case of reen- 
forced concrete beams, the writer believes that more tests are 
needed before such theorizing can be placed upon a permanent 
basis. Some results of tests of full-size reenforced concrete 
beams are given in the following tables : 



738 



APPLIED MECHANICS. 



SOME TESTS MADE IN THE LABORATORY OF APPLIED MECHANICS OP THE 
MASSACHUSETTS INSTITUTE OF TECHNOLOGY. 

Concrete was of the same composition as in the case of the columns. Size of beams 
8"Xi2". Span n'. When load was at two points they were 44" apart, and symmet- 
rical with reference to the center. 







No., Size, and Kind of 
Bars near Bottom. 


Manner of 


Weight 


Breaking 

Load, 
Exclusive 


Maximum 
Bending- 


No. of 


Age in 








Loading 


of 


of 
Weight 

of 
Beam, 
in Lbs. 


Moment 


Beam. 


Days. 


No. 


Side of, 

in 
Sq. Ins. 


Plain, P, 

or 
Twisted, 

T. 


at Time of 
Fracture. 


Beams 
in Lbs. 


at 
Fracture 

in 
In.-lbs. 


1 


40 








Center 


1 198 


1302 


62733 


2 


40 




i 


T 


' ' 


I20O 


1300 


62700 


3 


39 




§ 


T 


At two points 


1205 


10095 


241973 


4 


38 




1 

4 


T 


Center 


1 160 


13680 


470580 


5 


5o 




1 


T 


« 1 


1290 


14710 


506715 


6 


5o 




1 


T 


( < 


1204 


15796 


■S41134 


7 


4i 




ii 


T 


" 


JI 95 


12805 


442283 


8 


41 


2 


1 


T 


<( 


1240 


18760 


639540 


9 


42 


2 


il 


T 


< ( 


1274 


23105 


783486 


10* 


42 


2 


ii 


T 


< 1 


1279 


21105 


717569 


lit 


45 


2 


ri 


T 


tt 


1294 


23105 


783816 


12 


30 


2 


ii 


T 


At two points 


1282 


24200 


553553 


I3 x 


3i 


2 


ii 


T 


< < < « < < 


1292 


29200 


663718 


I4t 


30 


2 


ii 


T 


1 < 11 ti 


1341 


24200 


554527 


15 


53 


1 


I 


P 


it < < it 


1292 


I5 2 50 


356818 


16 


. 49 


1 


I 


T 


1 1 it tt 


1211 


16500 


382982 


17 


43 


2 


* 


P 


tt tt t 


1271 


15950 


370700 


18 


40 


2 


1 


T 


a tt tt 


1200 


19600 


438807 


19 


35 


4 


i 


P 


tt tt tt 


1261 


17500 


378065 


20 


33 


4 


i 


T 


it tt tt 


1213 


20000 


433329 


21 


57 


r 


J 


P 


t t t t t< 


1213 


12500 


295015 


22 


54 


2 


1 


T 


tt tt ti 


1248 


2*2250 


510092 


2 3 


57 


2 


1 


P 


it tt t> 


1221 


20250 


465647 


24 


47 


4 


t " 


T 


tt tt tt 


1203 


19250 


443350 


25 


5° 


4 


f 


P 


it tt a 


1 192 


152 50 


355i68 


26 


40 


2 


1 


T 


t t tt tt 


12 15 


24250 


553548 


27 


49 


I 


1 


T 


t 1 11 it 


1222 


21750 


498688 



* Also one bar \" square near top. 
t Also two bars \" square near top. 

In beams Nos. 12 and 13 there were, on each side of the middle of the span, 
eight pieces of J-inch twisted steel wire, bent in the form of a U enclosing the two 
reenforcing rods. In No. 13 they were vertical, and in No. 12 they were inclined 
at 45 to the horizon, sloping upward away from the middle. In No. 14 the 
wire pieces were in the form of a square, vertical, and enclosing all iour of the 
bars. Nos. 26 and 27, each contained, in addition to the bars, a vertical layer 
of expanded metal, extending throughout their length and height. 



HYDRAULIC CEMENTS AND BRICK PIERS. 



739 



SOME TESTS MADE AT THE ENGINEERING EXPERIMENT STATION, 
UNIVERSITY OF ILLINOIS. 

Plain Concrete. 
Beams were 12" wide by 13^" deep. Mixture by volume was 1 part Chicago A A 
Portland cement; 3 parts clean, sharp sand; 6 parts broken limestone (i"-ii.")» 



Beam No. 


Length. 


Age, days. 


Span. 


Maximum 
Applied Load. 


Modulus 
of Rupture. 




Ft. Ins. 




Ft. Ins. 






8 


15 4 


64 


14 


3600 


412 


11 


15 4 


65 


14 


2600 


337 


18 


15 4 


64 


14 


2400 


322 


26 


12 


62 


10 8 


550° 


39° 


3° 


12 


62 


10 8 


4800 


355 


23 


9 6 


61 


8 6 


6355 


347 


3 1 


9 6 


62 


8 6 


8000 


422 


24 


6 


61 


5 


10240 


2 99 


25 


6 


64 


5 


10200 


299 



Reenforced Concrete. 
Size of beams: Length, 15' 4", Span 14', breadth 12", depth i3t", center of metal 



reenforcement 12" below top surface of beam 
of beam 



Amount and Kind 


Area of 

Metal, 

Sq. Ins. 


1 

Maximum 
Load, 
Lbs. 


of Reenforcement. 


3 \" plain round 


•59 


9000 


3 r " " 


•59 


9200 


3 \" " square 


•75 


9900 


3h" " " 


•75 


I OOOO 


4 f " " " 


2.25 


26900 


3 \" Ransome 


•75 


22800 


3 I" Thatcher 


1.20 


18400 


3!" " 


1.20 


16600 


3 \" Kahn 


2 .40 


24400 


5 h" " 


2 .00 


23000 


4§" " 


1.60 


172OO 


3h" " 


1.20 


15000 


6 f " Johnson 


2 .19 


343°° 


7i" " 


1.40 


29000 


5 *" " 


1. 00 


20900 


5 r " 


1. 00 


20600 


3 \" " 


.60 


14000 


3f " 


.60 


14000 



Loads applied at the one-third points 
BRICKS AND BRICK PIERS. 

In this connection two 
sets of tests of brick piers, 
made at the Watertown Ar- 
senal, will be quoted here. 
The first is taken from 
Tests of Metals for 1886, 
and the second from Tests 
of Metals for 1904. 

The tabulation of the 
second series, which com- 
prises 26 piers, is given in 
the table on page 742. 



740 



APPLIED MECHANICS. 



The first series comprises 53 piers, in the construction of which two kinds of 
brick were used, viz., common hard-burned bricks and face-bricks, laid on bed, 
with joints broken every course. The tabulation follows: 



Face-Brick Pikrs. 



Weight 




Actual Dimensions. 




Ultimate Strength. 




















Cubic 










Area. 




Per 


Per 


Foot, 
lbs. 


Height. 


Cross-section. 




Total. 


Sq. In. 


Sq. Ft. 


ft. 


in. 


in. 


in. 


sq. in. 


lbs. 


lbs. 


tons. 


132.7 


2 


O. 


7.63 


7.61 


58.06 


141000 


2428 


174-81 


134-4 


2 


O.27 


7.64 


7.63 


58.29 


123400 


2117 


152.42 


130.2 


3 


11.95 


7.75 


7.68 


59-52 


122016 


2050 


147 60 


129.7 


4 


O. 


7.75 


7.70 


59-68 


I I 6000 


1944 


139-97 


127.6 


6 


0.37 


7.75 


7-75 


60.06 


117117 


I950 


140.4 


129.6 


6 


O.26 


7.85 


7-75 


60.84 


106470 


1750 


126.0 


125.2 


8 


O.56 


7.78 


7-75 


60.30 


102000 


1691 


121.75 




9 


II.27 


7.80 


7.70 


60.06 


IOO749 


1677 


120.77 


126.8 


10 


0-37 


7.82 


7.80 


61.00 


I IO5OO 


l8ll 


130.39 


126.4 


*5 


11.68 


II.60 


II.60 


134.56 


257300 


I912 


137.66 


129.0 


5 


11.27 


11.55 


II.50 


132.82 


258100 


1943 


139.89 


130.3 


5 


11. 


j n-55 
\ 4.20 


II.50) 

4.10 s 


115-61 


2I9659 


I900 


136.8 


129.5 


5 


10.09 


15-45 


15.40 


237.93 


499653 


2IOO 


151. 2 


126.0 


5 


11. 81 


15-45 


15.35 


237.16 


4687OO 


I976 


142.27 



Common-Brick Piers. 



120.8 
123-3 
125.4 
124.6 
121. 5 

123.3 
121. 4 
121. 4 

121. 

123. 1 

123.5 
125.8 

124.9 
125. 1 
123.2 
121. 7 
121.6 

720.8 

H9-5 



10.87 


7.65 


II. 13 


7.65 


0.37 


7.60 


11.62 


7.68 


1.62 


7.65 


1. 18 


7.60 


1.50 


7.65 


11.98 


7.60 


0.93 


7.60 


1. 31 


7.60 


11.06 


11.60 


10.75 


11.38 


11.58 


11.55 


II. 81 


11.40 


0.75 


11.45 


0.75 


11.48 


1.37 


11.45 


0.75 


11.50 


1. 00 


11.55 



7.52 

7.60 

7-55 
7-58 
7.60 
7-58 
7-63 
7-55 
7-55 
7-55 
11.40 

11-35 
11.50 
11.30 
11.45 
11.45 
11.40 
11.40 
11.45 



57-53 


161000 


2798 


201.45 


58.14 


157800 


2714 


195.40 


57-38 


111891 


1950 


140.40 


58.21 


101867 


1750 


126.00 


58.14 


144300 


2481 


178.63 


57-61 


132503 


2300 


165.60 


58.37 


90474 


1550 


in. 60 


57-38 


90800 


1582 


113.90 


57-38 


86070 


1500 


108.00 


57-38 


104200 


1815 


130.68 


132.24 


307800 


2327 


167.54 


129.16 


318500 


2466 


177-55 


132.83 


224100 


1687 


121.46 


128.82 


251199 


1950 


140.40 


131-10 


222870 


1700 


122.40 


i3*-45 


216200 


1644 


118.36 


130.53 


190700 


1461 


105.19 


131-10 


21 1 100 


1610 


115.92 


132.25 


178200 


1347 


96.98 



Core built of common brick. 



HYDRA ULIC CEMENTS AND BRICK PIERS. 



741 



Common-Brick Piers— Continued. 



! Weight 
per 
Cubic 
Foot. 


Actual Dimensions. 


Sectional 
Area. 


Ultimate Strength. 


Height. 


Cross-section. 


Total. 


Per 

Sq. In. 


Pei 

Sq. Ft. 


lbs. 


ft. in. 


in. 


in. 


sq. in. 


lbs. 


lbs. 


tons. 


126.2 


I II.OO 


( II.40 

I 4-55 


II 
4 


•40 ) 
-5o f 


109.48 


271500 


2480 


178.56 


127.7 


2 1. 12 


j H-45 
f 4-9° 


II 

4 


40 I 

55 ) 


IO8.23 


265400 


2452 


176.54 


127.3 


4 O.18 


j H-35 
I 4.7o 


11 
4 


35 1 
.70 ) 


IO6.73 


198100 


1856 


133.63 


127.7 


3 n.98 


i "-45 
( 4.80 


11 

4 


40/ 
■70 f 


107.97 


215200 


1993 


143-49 


118. 8 


6 0.75 


j H-35 
1 4.90 


4 


35 I 
60 j 


106 28 


162 100 


1525 


IIO.52 


124.3 
125. 1 


8 3.1? 
10 2.27 


( 11.50 

1 4.90 

( 11.50 

} 4.80 

15.50 

15-70 

15.50 

15.70 

j 15.45 

| 8.60 


11 

4 
11 

4 


50/ 

80 y 
40 1 

80 f 


108.73 
108.06 


184841 
157500 


1700 

1457 


122.40 
IO4.90 


123.0 
121. 8 
123.7 
120. 1 

124.3 


6 1. 18 
6 1.37 
9 1 1 . 00 
9 11.98 

6 0.68 


15 
15 
15 
15 

15 

8 


45 
65 
40 
60 

45 I 
30) 


239-48 
245-71 
238.70 
244.92 

167.32 


358200 
356600 
230200 
247400 

270100 


1495 

1451 

964 

IOIO 

1614 


107.64 

104.49 

69.41 

72.72 

Il6. 21 


125.6 


6 0.68 


j 15.50 
1 8.70 


15 

8 


45 I 
60 f 


164.66 


260200 


1580 


II3-76 


123.4 


9 11.25 


( 15.40 
} 8.40 


15 

8 


40 j. 
30 f 


167.44 


2I2I00 


1267 


91.22 


128.9 


9 10.56 


j 15.45 
{ 8.70 


15 

8 


40 I 
60 j" 


163. II 


202OOO 


1238 


89.14 


122.3 
125.0 
117. 1 
124.0 

124-5 


*I2 6.5 
fl2 6.5 
+ 4 0. 
t 3 n.25 
§6 1.1 


11.60 

11-55 

7.90 

8.00 

j n-55 

( 4-20 


11 
11 

7 
8 

11 
4 


60 
40 
90 
00 

55 I 
20 ) 


I34.56 

131.67 

62.41 

64.00 

115.76 


217200 

I93300 

7230O 

IO580O 

6080O 


1622 
1468 

1 158 

1654 
525 


II6.78 
IO5.69 

83-37 
II9.O9 

37-80 



* Laid with bond stones 4 feet apart. 
% Face-brick pier grouted. 



t Common-brick pier grouted. 

§ Face-brick pier, laid without mortar. 



The mortar was composed of Rosendale cement 1, sand 2. 
The piers were 21 months old when tested. In this series the 
mortar was kept purposely the same throughout, so that the 
variation in strength should be due to the variation in dimen- 
sions of the piers. The mortar, however, was found to be 
much stronger in some places than in others. 



742 



APPLIED MECHANICS. 



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FUNDAMENTAL PRINCIPLES. 743 



CHAPTER VIII. 
CONTINUOUS GIRDERS. 

§ 244. Fundamental Principles. — A continuous girder is 
one that is continuous over one or more supports ; i.e., one that 
has at least one support in addition to those at the ends. The 
principle of continuity is, that the neutral line is throughout a 
continuous curve over the supports, the tangent to one branch 
of the curve at the support being a prolongation of the tangent 
to the other branch. 

Whereas, in the girder supported at the ends, the bending- 
moment at the support is zero, in the continuous girder there 
is a bending-moment at the support, where the girder is con- 
tinuous. There is also a shearing-force at each side of the 
support, the sum of the shearing-forces on the two sides of 
any one support forming the supporting-force. 

In this chapter will be given the general methods of deter- 
mining the bending-moments, slopes, and deflections of con- 
tinuous girders. 

i°. When the loads are distributed. 

2°. When the loads are all concentrated. 

3 . When there are both distributed and concentrated loads. 

It is believed that the reader will thus have the means of 
solving all cases of continuous girders, and that, whenever it 
is desirable to have a set of simplified formulae for a small but 



744 APPLIED MECHANICS. 

definite number of spans, or for some special proportions or 
distribution of the load, he will be able to deduce such simpli- 
fied formulae from the more general ones. 

§ 245. Distributed Loads. — In this case we assume that 
all the loads are distributed, whether they are uniformly dis- 
tributed or not. The first step to be taken is, to find the bend- 
ing-moment over each support : this is done by using what is 
known as the "three-moment equation" which we shall now 
proceed to deduce ; and, in the course of the reasoning by which 
we deduce it, we shall derive a number of useful equations, ex- 
pressing bending-moment, shearing-force, slope, deflection, etc.,. 
at various points. 



Fig. 247. 

For the purpose in view, let us assume our origin at O 
(Fig. 247), and let 

M x = bending-moment at B, 

M 2 = bending-moment at O. 

M 3 == bending-moment at A. 

/, = OA. 

/_, = OB. 

F Q = shearing-force just to the right of O. 

F_ Q == shearing-force just to the left of O. 

F x = shearing-force at distance x to the right of origin. 

F_ x = shearing-force at distance x to the left of origin. 

Shear is taken as positive when the tendency is to slide th* 
part remote from the origin downwards. 

If S Q = supporting-force at O, 

S = F + /*_.. 



DISTRIBUTED LOADS. 745 

Beginning, now, by taking O as origin, and x positive to the 
right, — 

Let OC = x. 

CD = v = deflection at distance x from origin. 
w == load per unit of length (either constant, or vari- 
able with x). 
We shall then have, from the principles of the common 
theory of beams, 



\ = F -J* wdx; (1) 



i.e., the shearing- force at a distance x to the right of O is 
found by subtracting from the shearing-force just to the right 
of O the sum of the loads between the section at x and the 
support ; and this sum is 



1 



wdx. 



In a similar manner, if we were to take origin at O, and x 

positive to the left, we should have 



Li = F-o - \ X wdx. (a) 



In § 204 we found the equation 

— = F x 

dx 



■•• ~r — Fo — / wdx, 
dx J 



Hence, integrating between x = o and x = x, and observing, 
that, when x == o, M = M 2 , we have 



— M 2 = 7^* — / / wdx 2 , 



746 APPLIED MECHANICS. 

which reduces to 

M= Af t + Fox- f*f X w<fx>; (3) 

or, in words, — 

The bending-moment at a distance x to the right of is 
equal to the bending-moment over the support at the origin, 
plus the product of the shearing-force just to the right of the 
origin by the distance of the section from the origin, minus 
the sum of the moments of the loads between the section and 
the support about the section. 

Observe that this sum of the moments of the loads between 
the section and the support about the section has, for its math- 
ematical equivalent, the expression 



t/o *Jo 



wdx 2 ; 



and, as a particular instance, it may be noted, that when the 
load is uniformly distributed, and hence w is constant, this will 
reduce to 

WX 2 , . X 

= Iwx) -t 

2 2 

wx being the load between the section and the support, and - 

2 

being the leverage of its resultant. 

Now' write, for brevity, 



«/o 1/0 



wdx 2 = m; 

t/o «/o 

then 

M= M 2 -f Fox - m. (4) 

Now, from § 194, we have 

d 2 v __ M 
dx 2 ~ EJf 



DISTRIBUTED LOADS. 747 



Let ^ = slope at distance x to the right of the origin. 
a_j = slope at distance x to the left of the origin, 
c^ = value of a, when x = o. 
a_ e = value of a_ i when j = o. 
Then 

do C x M . , 

where f is an arbitrary constant, to be determined from the 
•conditions' of the problem. 

If, now, we substitute for M its value M 2 -j- F Q x — m, we 
shall have 



**—*-*J..*i + J, 'J. in -lit 



mdx 



To determine c, observe, that, when .a: =.o, a £ = (*<,; 

.*. c = tana 



dv , ,^ C x dx 

tan a, = — = tan Oo + M 2 I — 

dx ./_ EI 



. j-, C x xdx r x ma 

+ F "l El- Ik 



mdx , v 
EI- (5) 



Integrate again, and observe, that, when x = o, v .= o, and we 
obtain 

& = ^tanoo + M 2 C f — 

Jo Jo ^ 



743 APPLIED MECHANICS. 

Now write, for the sake of brevity, 

j wdx 2 , «, = J / wdx 2 , «„,= I I wdx 2 , 

o t/o t/o t/O t/ t/ 

-x'x* -x'Xi- "-rrs' 

' - X"X"^ '■ - X'X'# '- - jTjC% 

-xr'f^-x'x^ --=r'x"# 



the last four being derived by taking x positive to the left 
We shall have 

v — ^tana + M 2 n 4- F Q q — V; (7) 

and, if v x = deflection at A = vertical height of A above O, we 
shall have, by substituting l x for x in (7), 

v x = h tan a Q 4- M 2 n l 4- ^#1 — V x , 

Now, if we assume any horizontal datum line entirely below all 
the points of support, and let the height of B above this line be 
y b , that of A, f a , and thaf of O, y Q , etc., we shall have 

y a — y Q = •/ -in a Q 4- M 2 n t + i^, — K„ (8) 

And, if we put x = /, in (4), we shall have 

M z = M 2 + ^oA - **i 

_ J / 3 - J/ 2 + W l 
.% ^o = jt . (9) 

and, if we substitute this value of F in (8), we obtain, by redu- 
cing, 

y a - y Q = ^tanoo + J/ 2 («, _ &) +-^"+ 5j£ - F.j 



DISTRIBUTED LOADS. 



749 



and, solving for tan a , we obtain 



tan 



y a — y Q / Vi n\ q t m 1 g x V x , x 



This expression gives us the tangent of the slope at O in span 
OA ; and equation (9) gives us the shearing-force just to the right 
of O in span OA, in terms of M 2 , M v and known quantities. 

If we were to take the origin at O, as before, and x positive 
to the left instead of the right, we should have, in place of (4), 



in place of (9), 
and in place of (10), 

yb -jo 

tana_ = — j ■ 



M = M 2 + F_ Q x — m; 



(11) 
(12) 



+ M 2 



te - fe) 



M T 



+ 



V-* 



(13) 



But, since the girder is continuous, we must have the tangent at 
O to the left-hand part, a prolongation of the tangent at O to 
the right-hand part, as shown in Fig. 248. 
Hence we must have 



a_ c s= — a^ 
tana_ -f- tana c 



o. 



Hence, adding (10) and (13), we 
have 



y- - y° : y* -y° . M ( & «* , ?■ 




- m 



?i 



3 # 



- J/ T 



/_r 2 



/i 2 



;/; 






(14) 



and this is the " three-moment equation " for the case of a dis- 
tributed load, whether it be uniformly distributed or otherwise. 



75° AFPLIED MECHANICS. 



CASE WHEN SUPPORTS ARE ON THE SAME LEVEL. 

When the supports are all on the same level, then y a = y b 
y oy and the three-moment equation becomes 






/i 2 /_x 2 



^i^i m_ x q _ 1 V x V_. 

" "7F ~ "71?- + 7, + 717 = a (IS) 



MANNER OF USING THE THREE-MOMENT EQUATION- 

When the dimensions and load of the girder are known, all 
the quantities in the three-moment equation, whether we use 
(14) or (15), are known, except the three bending-moments, M„ 
M 2 , and M y 

Suppose, now, the girder to have any number of (say, seven) 
points of support ; then, by taking the origin at B (Fig. 247), 
we obtain one equation between the bending-moments at E, B, 
and O, the first of which, if E is an end support, is zero. Next 
take the origin at 0> and we obtain one equation between the 
three bending-moments at B, O, and A; and so, continuing, we 
obtain five equations between five unknown quantities. 

Solving these, we obtain the bending-moments over the 
supports ; and from these bending-moments, after they are 
found, we can obtain the shearing-forces, bending-moments, 
slopes, and deflections, by using the equations deduced in the 
course of the reasoning for the three-moment equation, as equa- 
tions (4), (5), (7), (9), and (10). 

SPECIAL CASE, 

when, the supports being all on the same level, the load on any 
one span is uniformly distributed over that span, and when the 
girder is of uniform section throughout. 



DISTRIBUTED LOADS. 75 I 

Let w x = load per unit of length on span OA, origin at O. 
w_ x = load per unit of length on span OB, origin at O. 
I = the constant moment of inertia of the section 

y a = yb = y* 
Then 



W \t \ 2 

m x = , 
2 


w_ x l_? 
m_ x — ; 
2 


1 2EI' 


*- - lEi' 


* I=Z 6E~f 




V _ W M 

1 — t~> r> 


r, = K '- / - 4 . 



24^/ 24^/ 

With these substitutions, the three-moment equation, either 
(14) or (15), becomes 

MJ_ X + 2M 2 (l_ x + I x ) 4- MJ X + JC^x/x 3 + w-xZ-x 3 ) = o. (16). 

This is a simpler form of the three-moment equation, applicable 
to this particular case only. 

Example I. — Suppose we have a continuous girder of 
uniform section, uni- 
formly loaded, and * a n - a 
of three equalspans, 

to find M B and M c , also the supporting-forces, shearing-forces, 
bending-moments, slopes, and deflections throughout. 

Solution. — Take the origin at B, and we have 

M x = 0, M 2 = M z = M h = Mci 

since 

4 = /-, - /, 
equation (16) gives 

S M h / + \wl* = o .\ M B = M c = -— . 



wl 2 wl 2 ivl 2 
IO IO 2 


wl 


/ 


2 



752 APPLIED MECHANICS. 



Next, to find the shearing-forces, we have, from (9), 



F_ r = 



equals shearing-force just to the right of B or left of C. 
Shearing-force just to the right of C or left of B = 

wl 2 wl 2 

10 2 „ . 
- = |«£ 

Hence supporting-forces are 

Bending-moment in span y4# at distance ^r from A, cr in 
span CD at distance „r from D, 

M = Iw/x - — . 
2 

Bending-moment in middle span at a distance ;r from i? or 

from C, 

% , wl 2 wlx wx z 

10 2 2 

Shearing-force in span AB or CD at a distance x from A 
or D, 

F = |#V — «o\ 

Shearing-force in middle span at distance x from B or 67, 

F = — ze/jt\ 
2 

Maximum bending-moment in span i?£ (when x = -), 

2 

__ _ Z£// 2 7#/ 2 7£'/ 2 _ Z£// 2 

10 4 8 40 



DISTRIBUTED LOADS. 753 



Maximum bending-moment in span AB or CD, 

2 5 5o 25 

Hence the greatest bending-moment to which the girder is 

wl 2 
subjected is that at B or C, and its amount is — . 

10 

Slope at B in middle span, from equation (10), 

wl 3 



wl 2 / I \ , wl 2 ( I \ v W/3 

tan ttB = -^-^ + — (^J - ^7 + 



_ w/y 1 1 \ wi 1 



•O 24/ I: 



E/\$0 60 24/ \2oET 

which denotes an upward slope at B towards the right. In the 
same way, the girder slopes upwards at C towards the left. 
The slopes at B and C in the end spans are, of course, down- 
wards. 

Slope in the middle span at a distance x from B, 

dv 1 ( wl 2 , wlx 2 wx 3 ) , 

tan a = — = —\ x H —\ -4- c. 

dx Ely 10 4 6 \ 

When x =. o, 

wl 3 wl 3 

tana = 4- ,\ c = 



120EI 120EI 



wit.- 



/3 , ^ l* X /X 2 _~X* 

20 10 4 6 

w 



120EI 



(/3 — I2/ 2 JC -f" 30IX 2 — 20.**) 



Deflection = v = i/ 3 x — 6/ 2 ** -f- 10/^ — c^), 

\20EI 



754 APPLIED MECHANICS. 

In order to make plain all methods of proceeding, the slope 
in the end spans will be found in two different ways, as 
follows : — 

For bending-moment, slope, and deflection in left-hand span 

at a distance x from B (or in the right-hand span at distance x 

from C), we have 

_, wl 2 , X . wx 2 

M= h ~wlx . 

10 5 2 

dv i ( wl 2 x , rwlx 2 wx*) 

tan a = — = — \ + * ~~\ + e. 

dx EI\ io io 6 ) 

When x = o, 

tan a = — - /. c = 



120EJ \20EI 



dv w [ / 3 / 2 x , x , , x*) 

.-. tan a = — = —\ + -^-Ix 2 - - \ 

dx EI{ 120 io io 6 ) 



w 



(— / 3 — I2l 2 X -f- $6/x 2 — 20.X 3 ) 



120EI 



Deflection = v = — (-A* - 6/ 2 ^ + 12/* 3 - 5^), 

120 EI 



We may, on the other hand, accomplish the same object by 
finding the slope and deflection in left-hand span at distance 
x from A, or in right-hand span at distance x from D, as 
follows : — 

tana = -*L = J, Hlnix _ *£}<& = * r |*£ - *U c. 
dx E/J (5 2 ] E7\s 6 J T 

When x = /, 

tana = 



50£/ 

W 3 a// 3 , «// 3 

-r c • • * — — 



120EI 3o£I - 40E/ 



DISTRIBUTED LOADS. 755 



tan 



_ dv_ __ w ( / 3 . lx 2 x*\ 
dx El\ 40 5 6 j 



12 



w { 

oEI\ 



— 3/3 4- 24/x 2 — 20JC 3 



a/ ( 



3/3* 4- 8/^3 — 5^1. 
The figure shows the mode of bending of the girder. 

Fig. 250. 

To find the greatest deflection in either span, put the ex- 
pression for the slope equal to zero, and find x by the ordinary- 
methods for solving an equation of the third degree, and then 
substitute this value in the expression for the deflection. 

Example II. — Continuous girder of two equal spans, sec- 
tion uniform, and load uniformly dis- 

tributed. A * * 

Solution. — Take origin at B. 

M t = M 3 = M A — M z = 0, M 2 — M B , /j = / 2 = /, w x = w 2 = w; 

therefore, from equation (16), 

4 MJ + \wl* = .-. M B == - ~ 

8 

Shearing-force either side of B = 



w/ 2 w/ 2 

+■ 

y 



F B = F_ B = 8 . 2 = \wL 



Supporting-force at B = ^wl. 
Supporting-force at A and C — \wL 
Shear at distance x from A or C, 

F = %w/ — wxo 



750 APPLIED MECHANICS. 

Bending-moment at distance x from A or C, 

., 2 . wx 2 
M= ±wlx — 

8 2 

Maximum bending-moment occurs when x = §/, 

M = -^ _ _2_^ = 2^!., 

64 128 128 

Hence greatest bending-moment to which the girder is sub- 

tSUL 2 

jected is that at B, and its magnitude is . 

8 

Slope at B y from equation (6), 



tana R == tan 



— _ Til ( l \ _ wl1 _i_ 
8 V~ zEl) 12EZ 



24E/ 

= ^M_L _ -I- + -Li =0, 
is/ (24 12 24 ) 

as was to be expected. 

Slope at distance x from A in span AB, 

tana = — \^-wlx 2 ] •+- ^. 

^/Vi6 6 / 

When x =z I, a = o ; 



48^/ 

,\ tan a = — = — - { -^-lx 2 T 

dx EI (16 6 48 



48EI 
Deflection, 

48^57 



(9/^ - 8*3 - /3) t 



For maximum deflection, we have 
16 6 



3 , , X* /3 

J /* 2 =s O 



0.44/. 



DISTRIBUTED LOADS. 757 

Maximum deflection = — —\ — 1 + 3 (0.44) 2 — 2(0.44)31 (0.44) 
4SEI { ) 

wl x 

= — 0.00^4 

^EI 

Example III. — In order to solve a case where no simplif- 
cations enter, on account of symmetry or otherwise, we will 
take a continuous girder of five spans (as shown in the figure), 
the spans varying in length from 3/ to 7/; the loads being 
uniformly distributed, and varying in intensity from *$w on the 
longest span to Jw on the shortest ; the beam being of uniform 
section. 

A 3/ B 4? C 5J D « E Tl F 

A Ivo A %w a 5«7 A \w A 3m? A 

Fig. 252. 

For this case we can use equation (16). 
Origin at i>, 

o -f 14/^/3 + A IM C + £[7^(27/3) + 6^(64/3)] = o, 
or 

$6M B + i6M c = — 573w£. 

Origin at C, 

4 /M B + i8/M c + 5 /^T D + >/3[6(6 4 ) + 5(125)] = °, 
or 

i6M B 4- 72J/ C 4- 2oM = —loogw/ 2 . 

Origin at D, 

5/M c 4- 22/M D 4- 6/M E + £[5(125) + 4(2i6)]w/3 = o, 
or 

2oi/ c 4- S8M D 4- 24^ = -1489W 2 . 

Origin at E, 

6/M D 4- 26/M E 4- —[4(216) 4- 3(343)] = o, 
4 
or 

24J/ D 4- lo^Afu, = —1893 a// 2 . 



758 APPLIED MECHANICS. 

The four equations are : 

$6M n -f- \6M C = — 573«'/ 2 . (x) 

i6M B 4 72J/" C -f 2oJ/ D = — 1009a'/ 2 . (2) 

2oJ/ c 4- 8SM D 4- 24J/ E = — 1489a// 2 . (3) 

24^/ D 4- io$M K = — 1893a// 2 . (4) 

Eliminate J/ E between (3) and (4), and we obtain 

i^oM c 4- 536^0 = -6839a// 2 . (5) 

Eliminate M D between (2) and (5), and we obtain 

2i44J/ B 4- 8ggSM c = — ioioiio// 2 . (6) 

Eliminate M c between (1) and (6), and we obtain 

234792/^= -1769839a'/ 2 . (7) 

M B = - 7-5379 w/2 > 

.*. from (1), J/ c = — 9.4299a// 2 , 

from (5), J/ D = — 10.4722a// 2 , 

from (4), J/ E = -15.7853a// 2 . 

Shearing-force just to the right of 

A = ~°- 7 ' 5379 + * l * wi = 7.9874^ 

3 

„ —9.4290 -I- 7. 5^79 + 48 , 
B = v ^ vy '^- —a// = 11.5270a//, 

-10.4722 4- 94299 4- 62.5 
c = a// = 12.2915a//, 

_ — 1 s-7^s 3 4- 10.4722 4- 72 , 
D = 6 wl = 1 1.1 145 w/, 

x- i5-7 8 53 + 73-5 , , 

£ = — «// ass 12.75500/^ 



DISTRIBUTED LOADS. 



759 



Sheari no-force to the left of 



■fc> 



j, 7-5379 + 3!-5 , , , 

B — wl = 13.0126a//. 

3 * 

i- -7-5379 + 94299 + 48 . , 

C = wl = 12.4730a//, 

4 

-9.429 9 + 10.4722 + 6 2.5 

Z? = wl = 12.7084a//, 

j? -104722 + 15.7853 + 72 QQ 

jC = 7 «// = 12.8855a//, 

-15-7853 + 735 , , 
wl = 8.2450a//. 



7 
Supporting-force at 

^ = 7.9874a//, C = 24.7645a//, -£" = 25.6405a//, 

/? = 24.5396a//, D = 23.8229a//, ^ = 8.2450a//. 

Shearing-force at distance ;r to the right of 

A in section AB = 7.9874a// — 70a:, 

i? in section BC — 11.5270a// — 6a/#, 

C in section CD = 12.2915a// — 50a:, 

Z> in section DE — 11.1145 a// — 40/^, 

.Zs in section EF = 12.7550a// — ywx. 

Bending-moment at distance x from 

7a/jc 2 



^4 in section AB = + 7.98740//* — 

/? in section ^C = — 7.5379a// 2 + 11.52700//*: — 
C in section CZ> = — 9.4299a// 2 + 12.2915a'/*: — 
D in section Z>is = —10.4722a// 2 -+- 11.11450//*: — 
E in section i?/^ = —15.7853a// 2 -}- 12.7550a//*- — 



2 ' 
dwx 1 

$WX 2 

4wx 2 
IT' 

Swx 2 



760 APPLIED MECHANICS. 

For the sections of maximum bending-moments (put shear- 
ing-force = o), — 

In AB, x = 1.1410/; 
In BC, x == 1. 92 11/; 
In CZ>, jc = 2.4583/; 
In Z>-£, .r — 2.7786// 
In is^, jc = 4.2517/. 

Hence the maximum bending-moments are respectively, 

in — 

Section AB, 



wx 



(—7 + 7-9874^= +4.55700*. 

Section BC, 

— 7-5319 w ? 2 + oax(i 1.5270/ — 3*) = 3.5347a*- 
Section £Z?, 

— 9.4299W/ 2 4- oa#( 1 2.2915/ — §#) = 5.6781a//*. 
Section /?is, 

— 10.4722W/ 2 4* 0/^(11.1145/— 2jv) = 4.96930//*, 
Section iTF, 

— 15.7853W/ 2 4- 0/^(12.7550 — f#) = 11.32970*. 



the right. 



tan 



Values of tan a = slope in every case in the span, towards 

__/?! «A ^"3^ m x q x V x 
tan« = Jf,^ _ -^ - — - — + -. 

Slope at B, 

w/ 3 /2 \ /2\mc 3 32a// 3 i6wl l 

». = -7-5379^- - -J + 9-4*99^ j^j - -37- + ^7- 

Z£// 3 
= 0.^^571 . 



DISTRIBUTE!} LOADS. 



761 



Slope at C, 
tana c = - 9 . 4299 g(| - f) + 10.472= r 



Slope at D> 



5W 3 625 f£//3 

625 W 3 ze// 3 



wl l wl z >]2Wl± 36W/ 3 

tana D = -10.4722-^(1 - 3) + 15.7853^7 - -^7- f ^7- 



= 0.7297 



tan 



Slope at ij, 

^l z (l 7\ 1029 a// 3 343 a// 3 
«- = - '5-7853^7^ - - y — — -El + 



EI' 
8 EI " - 6 '°* 26 El' 



The manner of bending, very much exaggerated, is shown in 
the accompanying figure. 



Fig. 253. 

Slope at A = —4.096 , slope at F — -(-23.4578 

EI ' ii 

For the deduction, see what follows. 
Slopes 111 General. 

Span AB, origin at A, 

w ( 7 ) 

tan a = El I 3-9937^ - i& \ + ^. 

When x = 3/, tan a = 0.3371^— ; 

• • ^(35-9433 - 3i-5) + c = 0.3371— 
.'. <r = —4.106 — - 

^ ( 7 ) 

.•„ tana = — ^ j 3.9937/^ — -* 3 — 4.106m- 



762 AT PL J ED MECHANICS. 

Span BC, origin at B, 

tana = 0.337.^ - 7-5379^7- + 5-7635^7 " ■ m . 

Span CD, origin at C, 

w ( 5 ) 

tana = -pj\ —1. 5983/3 — 9.4299/^ -h 6.1458/* 2 — 7*3?* 

Span Z>ii, origin at D, 
tana = Jj J 0.7297/3 — 10.4722/ 2 * + 5.55725/^ - -*si. 

Span isY 7 , origin at ij, 

tana = JU -6.0426/3 - 15.7853/** -f 6.3775/r 2 - ~^|. 

When * = 7/, 

tana = -—(-6.0426 — 110.4971 + 312.4975 - 172.5) 

Deflections. &1 

Span AB, 

w ( 7 I 

p = -^ j 1.3312/V3 _ — x* - 4.106/** >• 

Span BC, 

v = — J 0.3371/3,* — 3.7689/ 2 * 2 -f 1.9211/^3 - J... ■ 

Span CZ2, 

» = — ] —1.5983/3* — 4.7149/ 2 * 2 + 2.0486/^3 — x*l* 

Span Z^.fi", 

z> = j^-j 0.7297A*:- 5.236i/ 2 * 2 + 1.8524 + /*3 - _l. 

Span ZjT 7 , 

^ = ^L i _6.0426/3* - 7.8927/*** + 2.I258/*3 - ^1. 



CONTINUOUS GIRDER WITH CONCENTRATED LOADS. ?6$ 

t : : 

The maximum deflections can be obtained by putting the 
slopes equal to zero, as before. 

§ 246. Continuous Girder with Concentrated Loads. — 
For our next general case, we will 
take that where there are no dis- —, ^2 T *? ■ A 



tributed loads, but where all the n "** < >< > n + 1 

loads are concentrated at single } " 

points, and the section uniform Wn "l, Wn 

r ' Fig. 254. 

throughout ; and we will begin by- 
assuming only one concentrated load on each span. 

Let the support marked n — 1 be the in — i) th support, and 
the length of the (n — i) th span be i_ x ; let the load on this 
span be W n _ „ and likewise for the other spans. Assume the 
origin at n, and let 

F n = shearing-force just to the right of n. 
F_ n = shearing-force just to the left of n. 
Fj = shearing-force at distance x to the right of n. 
F_ j = shearing-force at distance x to the left of n. 
Shear is taken as positive when the tendency is to slide the 
part remote from the origin downwards. 
If S„ = supporting-force at n> 

S n = F„ + F_ n . (1) 

Let, also, x n = distance from origin to point of application 
of load W n , and let x n _ x = distance from origin to point of 
application of load W n -x- 

Take x positive to the right. Then, for 



x < x„, F t = 
x > x n , F, 



' I = Fn > I (2) 



Moreover, we have 



dM -.. 

'dx~ = Fl ' 



764 APPLIED MECHANICS. 

hence, by integration, for 

dM = I F n dx; 

M K Jo 

Jf*M f*x f*x 

dM = / F n dx - / W„dx + c; 
M n Jo Jo 

the value of c being determined from the condition, that, when 
x = x n the two results must be identical. Hence we have, for 

x<x n , M=M n + F n x; ) .v 

x >x n , M= M n + F n x - W„(x - x„). \ U 

Make x = l n in the last equation, and we have 

M H + 1 = M n + FJ n - W n (l n - Xn). (4) 

Now let l n — x n = #„, and (4) becopies 

J^ + i = ^f« 4- ^«4 - W n a n ; (5) 



hence 



/? = ^i-"^+ ^* g * ($\ 



Moreover, we have, as before, 

d 2 v M 7?rd 2v 1* 

dx 2 EI dx 2 

I being a constant. 
Let, as before, — - 

a x = slope at distance x to the right of origin. 
a_ x r= slope at distance x to the left of origin. 
a n = value of a x when x — o. 
a_ w .= value of a_, when x = o. 



CONTINUOUS GIRDER WITH CONCENTRATED LOADS. 765 

Then by integration, determining the constant in the same 

way as in (3), we have, for 

x 2 

x < x„, Ef (tan a, — tan a„) = M„x + F n — ; 

x > x„, ^/(tano, - tana„) = M n x + F n - - W *(* ~ *")\ 

2 2 

Hence 

di) x 2 

x < x n , EI-~ = £7 tana* + M n x + /^— ; 
dx 2 

x > x„, EI^L = Elxz&on + M n x + ^«- - W *(* "" *»>* . 
tfjc 22 

Integrate again, and determine constants in the same way, 

and for 

x 2 x 3 

x < x„, EIv — EIx tan a* -f M n h F n — ; 



x > x n , EIv = EIxtena n + M n - + /?„- - ^ ( * - x «)\ 

26 6 



(8) 



Make x = 4 in the last equation, and denote the heights 
of the supports above the datum line in the same way as in 
§ 245, and we have 

EJUn + 1 — yn) = EIl n tan On 

+ mJI + f!± - W *V" - *< ( 9 ) 

26 6 

Substitute for l n — ;tr„, <z w , and for F ni its value from (6), and we 

have 

^(7* + 1— y*t) = EIl n tan a„ 

+ Jlfj*. +M n+ ^+ ^Vn* - *„*)• (IO) 

3 o o 

Hence 

Eltona* + ^( 2 J/„+ J/„ + I ) + -^(4* - * n *) 
6 64 

At 



j66 APPLIED MECHANICS. 

Now, if we take origin at n and x positive to the left, we 
should obtain, instead of (n), 

EI taxi a_„ + %i(2if„ + -%_«) 
6 

+ y ;;^- (4-.' - *„_.•) = *ff*=^M (. 2 ) 

6** — i \ 4 _ i / 

Now add (n) and (12), and observe, that, since the girder is 
continuous, 

tan an 4- tan a_« = o, 

and we obtain 

^(4 + 4-x) + Mn_kf± + *iJ* 
3 6 6 

«^{ * + > -■*■ + *-'-* }. (13) 

I 4 4 — I J 



and this is the "three-moment equation" for the case of a 
single concentrated load on each span, and a uniform section. 
When the supports are all on the same level, this becomes 

^(4 + 4.,) + M n J±f± + *»±£ + -^(4 2 - V) 

3 6 6 64 

+ ^-^-' (4,,* - *„_,*) = o. (14) 

64 _ 1 

Either of these equations can be used (when it is appli- 
cable) just as the three-moment equation was used in the case 
of distributed loads. 



CASE OF MORE THAN ONE LOAD ON EACH SPAN. ?6j 



CASE OF MORE THAN ONE LOAD ON EACH SPAN. 

When there is more than one load on each span, the three- 
moment equation becomes as follows : — 

^(4 + 4-0 + M n _ *=-« + M n+ £ 
3 6 6 

64 64 _ j 



( 4 4 — t ) 



In using these equations for concentrated loads, we can 
determine the moments over the supports ; but we must observe, 
that, in getting slopes and deflections, bending-moments, etc., 
the algebraic expressions that represent them are different on 
the two sides of any one load, and hence we must deduce new 
values whenever we pass a load, determining the constants for 
our integration to correspond- 

Example. — Given a continuous girder of three spans, the 
middle span = 20 feet, each end span =15 feet ; supports on 
same level. The only loads on the girder are two ; viz., a load 
of 5000 lbs. at 5 feet from the left-hand end, and one of 4000 
lbs. 5 feet from the right-hand end. The supports are lettered 
from left to right, A, B, C, D, respectively. Find the greatest 
bending-moment and greatest deflection. 

Solution. — Origin at B, 

^( 2 o + , S ) + ^( 2 o) + S^_l (MS -, 5 ). 8u (I) 
Origin at C, 

^(,o + , S )+^(*o)+<J^(»5-.5)-o. (.) 



768 



APPLIED MECHANICS. 



These reduce to 






3 


= 


M B 


= —4000 foot-lbs. 


3 


= 


• 
• • 


=5 —2667 foot-lbs. 


Shearing-forces. 


Supporting-forces. 


Slopes at supports. 


Fa = 3067, ^-c= -67, 


S A = 3067. 


tan a A - - 59444 


-^-b = i933> Fc = 15"- 


«S B = 2000. 


tana„ = + 35556. 
EI 


F* m 67, /L D = 2489. 


S c = 1444- 


31111 




S -- 


s 2489. 


tan «_„=-» 
EI 



Span ^2?, origin at ^4, 

* < 5, M = 3067*. 

x > 5, J/= 3067* - 5ooof> - 5) = 25000 - 1933*- 

Maximum bending-moment occurs when x = 5 and therefore 

M = 15333. 

* < 5, EI tan a, = -59444 + I533-* 2 J 
jc > 5, EI tan a x = 25000* — 967J? 2 •+• c. 

Determine c by condition, that, when x = 5, these two become 
equal; 

.-. c = -121944; 

/. x > 5, EI tan a, == — 121944 -J- 2500a? — 967**, 

For deflections, 

* < 5, .£/# = - 59444* + 5 IIx3 '> 

x > 5, 2?/# = — 121944.2; + 1250a* 2 — 322a: 3 -f- *. 

Determine c from condition, that, when x = 15, z> = o; 
.\ *• = 104167 ; 

•*• * > 5> 2*/z> = 104167 — 121944* -f 12500** — 322*'. 



CASE OF MORE THAN ONE LOAD ON EACH SPAN. 769 

For maximum deflection, equate slope to zero, and find x. 

We find it at x = 6.53. 
.-. EIv Q = —249531. 

Span BC, origin at B, 

M — —4000 + 674:, 
EI taxi a, = 35557 — 4000* + 33**, 
EIv — 35557.* — 2000.x 2 + IMP*. 

For maximum deflection, equate slope to zero, and find X, 

We find it at x = 9.78, 
.\ EIv — it 6^40. 

Span CD, origin at C, 

x < 10, M = —2667 + i$iix; 

x> 10, M = —2667 + 1511^ — 400Q(> — 10) =s 37333 — 2489*; 

x < 10, EI tan a r = —31 in — 2667.2: + 756^ ; 
x> 10, EI tan a x = —23 mi -f- 37333* — 1245.*?. 

For deflections, 

x< 10, EIv = — 31111* — 1334* 2 + 252a- 3 ; 

x > 10, EIv = —231 1 1 1* + 18667.x: 2 -- 415.x: 3 ■+• c. 

When x = 15, v = o ; 

.*. x > 10, EIv = —37132785 «— 2311113:-}- 18^67^ — 4 v$x*. 

For maximum deflection, equate slope to zero, and find x. 

We find it at x = 8.41. 
•\ EIv = —24506. 

Hence greatest bending-moment and greatest deflection are 
both in span AB. 

Observe, that, since we have used one foot as our unit of 
measure, all dimensions must be taken in feet, and the value 
of E is also 144 times that ordinarily given. 



77° APPLIED MECHANICS. 

§ 247. Continuous Girder, with both Distributed and 
Concentrated Loads. — In this case we may either calculate 
the bending-moments, slopes, and deflections due to each sepa- 
rately, and then add the results with their proper signs, or we 
may modify the solution that was used for the case of a dis- 
tributed load, so as to extend its applicability to this case. 

Let W represent any one concentrated load, and let x x rep- 
resent the distance of its point of application from the origin. 
Then, in the general formulae deduced for the distributed load, 
make the following changes ; viz., — 

1°. Instead of 

wdx* 9 

■ 

put 

* wdx* + ^JV(x-x t ) 9 



*Jo «/o 



since, as was shown, m represents the sum of the moments of 
the loads, between the section and the support, about the sec- 
tion. 

2°. Instead of 

Jf x C* 1 t%x ° A ) dx 2 

o Jo I Jo Jo ) EP 

put 

and make the corresponding changes in the values of m„ *#_,, 
V It and F_„ leaving n and q just as before ; then use the same 
three-moment equation as before, with these substitutions, i.e., 

M iii _ S + fci _ 5-1 _ jf s ?i - jf*=i -'Mi- **fe 



SPECIAL CASE. Jj\ 



SPECIAL CASE. 

when the distributed load is uniformly distributed on each span, 
but may be different on the different spans, and when the girder 
is of uniform section. 

Let w x = weight per unit of length on OA. 
w_ t =■ weight per unit of length on OB. 

Denote by W t any concentrated load on OA at distance x x 
from O. 

Denote by W_ x any concentrated load on OB at distance 
x_ x from O. 

Then we shall have 

z 

and, as before, 

/x 2 /_,■ 

Making these substitutions in the three-moment equation, 
and clearing fractions, we obtain for the case, when the sup- 
ports are all on the same level, 

o = M X L X + 2M 2 {h + /_,) + M z h + \(wj? + W-xLft 

+ 7^1 W,\l* - (/, - x x y\{l x - x x )\ 

•I 

+ j- a I w- ,[/.,' - (/_, - *-.)»](*-, - *-.)*• 



772 APPLIED MECHANICS, 



CONCENTRATED AND DISTRIBUTED LOADS. 

Example. — Let the girder be of uniform section, of two 
equal spans, each being 10 feet ; let the concentrated loads be 
A D ^ as shown in the 



4' ~x 3' J< 4' >[< 3 ' > figure, the dis- 
* tributed load be- 

1000 1000 

ing 96 lbs. per 
2000 FlG - 255 ' foot. Find the 

value of EI, so that the deflection may nowhere exceed ^q of 
the span. 

Solution. — Use equation (12); and, in deducing value of 
M B , use dimensions in feet ; afterwards use inches. 

Originate M A = M c = o; 

40 M B -f - 9 4 6 -(iooo + 1000) 

+ £j\ 2000(64) (6) + 1000(51X7) -f- iooo( 9 i)(3)f = o, 

4oM B -f 48000 -f 139800 = o, or M B = —4695 foot-lbs., 

M B — —56340 inch-lbs., M K = M c = o. 
m x = 177600, m_ t = 201600, 





7200 


n x 60 


7200 


«_i 60 


«I 


= -£1' 


X = El' 


*- = EI' 


IZ^EI' 




288000 


Qi 20 


288000 


q~x 20 


ft 


" EI ' 


/ x 2 ~~ EI 


*-* = ~eT> 


/_ x 2 " EI 


v x 


I75680000 

" EI ' 


V x 1464000 193536000 
/, ' EI - 1 " EI ' 


V_ x 161 2800 
/_, " EI 




Shear right side of middle 


+ 56340 -j- 177600 
= — 1949.5, 

I20 W D> 




Shear left side of middle 


O -f 5634O -|- 20I600 
= = 2149.5 9 
I20 W '' 




Shear left end 




--5634O + I53600 

as ■ 

I20 


— 810.5, 




Shear right end 


— 5 6 34<> + 177600 
120 


as IOIO.5; 




Middle supporting-force 


— 4099. 





slopes. 773 



Bending- Moments in Each Span. 
Span A B y origin at A, 

810.5.* — 4* 2 ' 
or 

810.5.* ~~ 4* 2 — 20oo(# — 72). 

Span BC, origin at B, 

-56340 + 1949-5* - **> 

— 56340 + 1949-5* - 4* 2 - iooo(* — 36), 

— 56340 -h 1949. $x — 4X 2 — iooo(jv — 36) — iooo(jp — 84). 

To ascertain position of the greatest bending-moment, dif- 
ferentiate each one. 

810.5 — 8x =5 o, jc = 101.31 ; 

810.5 — 8x — 2000 =0, x = a minus quantity; 

1949.5 — 8x = o, x = 243.69 ; 

1949.5 — 8x — 1000 = o, x = 118.69; 

1949.5 — 8x — 1000 — 1000 = 0, x = a minus quantity. 

Hence, in span AB, maximum bending is at the load, and 
its amount is 

(810.5X72) - 4(72X72) = 37620. 

Span BC, maximum is at right-hand load, and is 

-56340 + 1949-5 ( 8 4) - 4(84) (84) - 1000(48) = 31194. 



SLOPES. 

Slope at B y 

-56340 (177600) (20) 1464000 165600 
T = -EI-( 2 ° " 6 °) - " EI + —EI- = ~ET- 

Slope and Deflection in Span AB. 
First part, 

tana = ^1405.25^ — -**£ + tanao, 
a,, being slope at A. 



774 APPLIED MECHANICS. 

Second part, 

1 ( 4 ) 

tana = -pj\ 405.25.2? * 3 . — iooa* 2 + 144000* > -f c* 

When x = 72, a is the same in both cases ; 



J-jiooo(72) 2 - (144000) (72) i + tanoo - c « o 

5184000" 
c — tancto — — -=rj — • 

165600 



When x = 120, the second value of tana becomes 



EI ' 



•". £j\ (4O5.25) (l20) 2 --(l20)3- (IOOO) (I20) 2 + I 44 000(l20) i 



5184OOO 16560O 

+ tana ___ = __^ 

tanao = -M-6411600 -f 5184000 + 165600J = - IQ ^ Q0 , 
6246000 

Hence slope in first part (between A and the load), 

tana = -pj\ —1062000 + 405. 25** — -* 3 J* 

Second part (between B and the load), 

tana = -gj\ —6246000 + 144000* — 594-75** ~ \ x% \* 

Deflection. 
First part, 

v = -^-\ —1062000* -f 135.08*2 — —I. 
EI\ *' 3) 

Second part, 

9 = "k~ T \ —6246000* + 72000** — 198.25** — -*n +§• 

EI ( 3 J 



slopes. 775 



When x = 1 20, z; = o ; 
1 ( (120) 4 ) 

-^jj (62460OO) (I20)- (72000) (l20) 2 + (198.25) (l20)H " \ 

= ^(1036800) = "44i6oQQ a 
£7 V 6 ' EI 

Point of greatest deflection is found by putting slope equal 
zero. Moreover, it is plain that the greatest deflection is in the 
first, and not the second, part. 

Hence equation is 

■|^ 3 — 405. 25.x 2 + 1062000 = o 
.-. *= 56^.77; 

and, substituting this in the expression for the deflection, we 
obtain 

39037720 



v a = — 



EI 



.120 39037720 , . 

Hence, putting — = ^7 — , we obtain 

r & 400 EI 

EI= 130125733. 

If E = 1400000, / = 92.9 ; therefore, if b = 3 inches, h = 
7 inches. 

Slope and Deflection in Span BC. 
Portion nearest B } 

tana = -=j J 165600 — 56340* -I- 974.8JC 2 x*y 

When x = 36 inches, we obtain 

1 661507 
tana = -^(165600 — 2028240 -f 1263341 — 62208) = Ff • 

Middle portion, 

tena = El\ - 20 34<Wf + 474-75* 2 — -**\ + *• 



77° APPLIED MECHANICS. 

When x = 36 inches, then tan a = — =r== — ; 

111 

.\ —661507 = —732240 + 615276 — 62208 -I- EIc 

... A 482335 

•' El 

.\ tana = -gy j -482335 — 20340* + 474-75** * 3 |* 

When x = 84 inches, 

tana= -^(--482335 - 1708560 + 3349836 — 790272) 

368669 
"* + EI 
Portion nearest C, 

i = •^<6366o.* — 25^ x*\ -f c. 

When x = 84 inches, then tan a = — ^r^- \ 

.*. 368669 = 5347440 — 176400 — 790272 + -£/<: 

4012099 
•'* c = ZT* 

.*. tana = pTjX —4012099 + 63660* — 25a? 2 x 3 >* 

When x = 120 inches, 

1 , , v 963101 

tana = -^(—4012099 + 7639200 — 360000 — 2304000) = — p~T~* 

Deflection. 

Portion nearest B, 

v — ~-| 165600* — 281 70* 2 + 324.9*3 — I.# 4 l. 

When x = 36 inches, 

v = -£-(165600 — 1014120 + 421070 - 15552) (36) 

(443002)36 15948072 

EI " EI 



SLOPES. 



777 



Middle portion, 

v = -£j\ -482335* - 10170* 2 -f 158.25*3 _ 1*4 1 4- c. 

15948072 
When x = 36 inches, then v = ^ — ; 

.% -15948072 = (-482335 - 366120 4- 205092 - 15552) + EIc, 

15289157 



c = 



^/ 



,\ z/ = -^-j -15289157 -482335*- io^o^-f 158.25*3— I*4|. 

Greatest deflection occurs in the middle portion, and the 
point is given by the equation. 

o = -482335 - 20340* + 474.75** - i* 3 = o; 
/. * = 71.4. 
Greatest deflection in span BC f 

a B^= =^ __^ --si^-S 

Fig. 256. 
V = 

^(-15289157-34438719-51846253+57602105-8662899) 

52634923 



^/ 



120 52634923 , . 
Hence, putting — — = =r^— , we obtain 

4OO tLl 

£/= 175449743; 
therefore, if E = 1400000, we have 

/= I2 5-3- 
If b = 3 inches, h = 8 inches. 



77$ APPLIED MECHANICS. 



EXAMPLES OF CONTINUOUS GIRDERS. 

i°. Let / = uniform moment of inertia of girder. 

w = load per unit of length uniformly distributed. 
Find expressions for 

1, the bending-moment over each support, 

2, the supporting-forces, , 

3, the greatest bending-moment, 

4, the slopes at the supports, 

5, the greatest deflection, 

in each of the following cases : — 
(a) Two equal spans, length /. 
{b) Three equal spans, length /. 

(c) Four equal spans, length /. 

(d) Two spans, lengths / x and l 2 respectively. 

(e) Three spans, lengths /„ l 2 , and / 3 respectively. 
(/) Four spans, lengths /„ l 2 , / 3 , and / 4 respectively. 

(g) Two equal spans ; loads per unit of length on each span, 
w 1 and w 2 respectively. 

(k) Three equal spans ; loads per unit of length on each span, 
w I} w 2 , and w z respectively. 

2°. Do the same in the case where each span is loaded with 
a centre load W, and has no distributed load. 

3°. Find greatest bending-moment and greatest deflection 
for a continuous girder of two spans, uniformly loaded on these 
two spans with load w per unit of length, and which overhang 
the outer supports ; the overhanging parts having lengths /_ 
and 4 respectively, and the same distributed load per unit of 
length on the overhanging parts. 



EQUILIBRIUM CURVES. — ARCHES AND DOMES. 779 



CHAPTER IX. 



EQUILIBRIUM CURVES. — ARCHES AND DOMES. 



§ 248. Loaded Chain or Cord. — It has been already shown 
(§ 126), when the form of a polygonal frame is given, that the 
loads must be adapted, in direction and magnitude, to that form, 
or else the frame will not be stable. The same is true of a 
loaded chain or cord, which would be realized if the frame were 
inverted. 

If a set of loads be applied which are not consistent with 
the equilibrium of the frame under that form, it will change its 
shape until it assumes a form which is in equilibrium under the 
applied loads. 

As to the manner of finding (when a sufficient number of 
conditions are given) the stresses 
in the different members, etc., this 
was sufficiently explained under the 
head of " Frames," and will not be 
repeated here, as the figures speak 
for themselves. 

In Fig. 257 the polygon fedcbaf 
is the force polygon, while the equilibrium polygon is 123456, 
an open polygon. A straight line joining e and a would repre- 
sent the resultant of the loads. 




Fig. 257. 



780 



APPLIED MECHANICS. 



CHAIN WITH VERTICAL LOADS. 

If all the loads are vertical, the broken line edcba becomes 

a straight line and vertical, as 
shown in Fig. 258^. Whenever the 
loads are concentrated at single 
points, as 2, 3, 4, 5, the form of 
the chain is polygonal ; and when 
the load is distributed, it becomes 
a curve, as shown in Fig. 259. 




Fig. 258. 




CURVED CHAIN WITH A VERTICAL DISTRIBUTED LOAD. 

Given the form of the chain AOE supported at A and E, 
and the total load 
upon it (be, Fig. 
259b), to find the 
distribution of the 
load graphically. 
First lay off be to 
scale, to represent 
the total load : this 
is balanced by the two supporting-forces at A and E respec- 
tively, as shown in the figure. Hence draw ca parallel to the 
tangent at E, and ba parallel to that at A, and we have the 
force polygon abca; the equilibrium curve being the chain AOE 
itself. Moreover, if the lowest point of the chain be (9, then 
the load must be so distributed that the portion between O and 
A shall be balanced by the tension at O and that at A, and 
hence that its resultant shall pass through the intersection of 
the tangents at O and A. Its amount will be found by drawing 
from a a horizontal line ; and then we shall have ao as the ten- 
sion at o, ab as the tension at A, and bo as the load between A 
and O. Hence the load between E and O will be oc. 



LOADED CHAIN OR CORD. 78 1 

Moreover, the load between O and any point, as B, will be 
balanced by the tension at O, and the tension at B, and hence 
will be od, where ad is drawn parallel to the tangent BD, so 
that the load between B and E will be dc ; and in this way we 
see that we can find the tension at any point of the chain by 
simply drawing a line from a, parallel to the tangent at that 
point, till it meets the load-line be. 

It is to be observed, that, if the tension at any point of the 
chain be resolved into horizontal and vertical components, the 
horizontal component will, when the loads are all vertical, be a 
constant, and the vertical component will be equal to the por- 
tion of the load between the lowest point and the point in 
question. 

If we assume our origin at O, axis of x horizontal and axis 
of y vertical, and let the co-ordinates of B be x and y, and if w 
be the intensity of the load at the point (x, y), we shall have, for 
the load od between O and B, 

P= J^wdx; 
and, since the angle oad = angle BBC, we shall have 







dx 


_ BC 
DC' 


od 
oa 


P 
H 




By 


differentiation, we shall have 










d 2 y _ 
dx 2 ~ 


\dx ) 
H 




or 






d 2 y 

dx 2 


w 


* 



(•) 



and this is the equation for all vertically loaded cords. 



782 



APPLIED MECHANICS. 




From it we can find the form of the cord to suit a given 
distribution of the load. 

§ 249. Chain with the Load Uni- 
formly Distributed Horizontally. — 
In this case w is a constant ; and if 
we assume our origin at the lowest 
FlG - 26 °- point of the chain, and use the same 

notation as before, we shall have 

d 2 y _ w 

dx~ 2 ~ w 

Hence, integrating, and observing, that, when x = o, -^ = O, 
we have 



dx 



wx 



and by another integration, observing, that, when x = o,y = 0, 

we obtain 

w 



y = 



2H 



■X 2 . 



This is the equation of a parabola; hence a chain so loaded 
assumes a parabolic form. 

Example I. — Given the heights of the piers for support- 
ing a chain so loaded, above the lowest point of the chain, as 
8 and 18 feet respectively, the span being 100 feet, to find the 
distance of the lowest point from the foot of each pier, and 
the equation of the curve assumed by the chain. 

Solution. — If (with the lowest point of the chain as origin) 
we call (x lt y^ the co-ordinates of the top of the first pier, and 
(x 2i y 2 ) those of the top of the second pier, we shall have, since 
y x = 18 and^ 2 = 8, and since we must have 



w , 



CHAIN WITH UNIFORM HORIZONTAL LOAD. 783 



8 = ^ 





w 
18 = S* 


and 


••• - = \ 


As 3 

' 8 ~ 2 


# X = -X, 

2 


Xt + x 2 = 


100 .*. f.# 2 


= IOO 



#1 -f- #2 #2 > 



# 2 = 4O, ^ = 60. 



Hence, since 18 = ~7}(^°Y 



2H 3600 200* 

therefore equation of the curve is 

y = -sh**- 

Example II. — Given the load on the above chain as 4000 
lbs. per foot of horizontal length, to find the tension at the low- 
est point, also that at each end. 

Solution. 

w 1 

-— = — , w = 4000, 

2H 200 

.*. 2H = 800000 ,\ H = 400000 lbs. 

Moreover, load between lowest point and highest pier £= 
60 X 4000 = 240000 lbs. 

Therefore tension at highest pier = 

y/(24000o) 2 + (400000) 2 = iooooy/(24) 2 + (40) 2 

= iooooy/2176 = 466480 lbs. 
Tension at lowest pier = 



y( 160000) 2 + (400000) 2 = iooooy/256 -f- 1600 

= iooooy/1856 = 430813 lbs. 



784 APPLIED MECHANICS. 

Example III. — Given the span of the chain as 20 feet, and 
its length as 25 feet, the two points of support being on the 
same level, to find the position of the lowest point. 

§ 250. Catenary. — The catenary is the form of the curve 
of a chain, which, being of uniform section, is loaded with its 
own weight only, i.e., with a load uniformly distributed along 
the length of the chain. 

To deduce the equation. of the catenary: if we assume the 
origin, as before, at the lowest point of the curve, we shall 
have still the general equation 

d 2 y _ w 
dx~*~ H' 

but w in this case is not constant. 

If we let w x = the load per unit of length of chain, we 
shall have 





w = 


Wi i = "V 1 + (H 




hence 




d 2 y w t ds 
dx 2 H dx 




Or, if we let 




w x 1 
~H~ m 




a constant, 




d 2 y 1 ds 
dx 2 m dx* 





(«> 

which is the differential equation of the catenary ; and we only 
need to integrate it to obtain the equation itself. 
To do this, we have 

d*y 
d z y t / ///1A2 dx 2 



my ' \dx) 






l/ 



dx/ 



THE CATENARY. 



78S 



therefore, integrating/and observing, that, when x = o, -~ = o, 

ax 

we shall have 



Mi-Mm- 



dy if £ -*\ ml * , -*\ 

... -£-J~-*-) A ,= -(- + <*') + <. 

But, when .r = o, ^ = O .*. £ = — m; hence the equation is 



m/ 

= 2\ 






(3) 



and this is the equation of the catenary 
when the origin is taken at O, the low- 
est point of the chain. 

If it be transferred to O lf where 00 1 
^z m, the equation becomes (by putting 
for y,y — m) 







(4) 



This is the most common form of the equation to the cate- 
nary, the origin being taken, at a distance below the lowest 

TJ 

point of the curve equal to tn = — , the horizontal tension 



w, 



divided by the weight per unit of length of chain. 



786 . APPLIED MECHANICS. 



To find x in terms of y, 


we have 


X 

e m -f 


1 ._ 2 y 


?* . 2V * 

m 


*x 2V x 

m 


Solving, we have 




*£ = L ± J^_ _ i 

m V m 2 


... f-iog,iz :t i/4_ I i 



(w t m 2 ) 



(s) 



To find the length of the rope : from the equation 

m( £ , -*\ 
y = —\e™ + e m ) 

we obtain 

dy i/ * _£\ 

dx 2\ / 



.\ s = - I \f™ + '~^V* =—(J* — *"£). (7) 

To find the area OO x A^B, we have 
Area = I ydx = — / Uk -f e~™)dx = — (** — *~*Y (8) 



TRANSFORMED CATENARY. ffl 



But 

arc OB = —(e^ — e~»n ■■ // 



hence area OO.A.B =. ms. 

This shows, that, if the load should be distributed in such a 
way as to be like a uniformly thick sheet of metal, having for 
one side the catenary and for the other the straight line O x A„ 
the equilibrium curve would be a catenary. 

It may be convenient to have the development of e m and 

e m ; hence they will be written here : — 

A = x + £ + -£L + _£L + J£- + etc., 
m m 2 \z m 3 ^ m 4 ^ 

m /# 2 [2 w 3 j3 z« 4 ^ 

Example I. — Given a rope 90 feet long, spanning a hori- 
zontal distance of 75 feet ; find the equation of the catenary, 
the sag of the rope, and the inclination of the rope at each 
support, supposing these to be on the same level. 

§251. Transformed Catenary. — We have just seen that 
the catenary is the form of chain suited to a load which may 
be represented by a uniformly thick sheet of metal, with a hori- 
zontal extrados, provided the distance 00 1 is equal to m> a defi- 
nite quantity. A more general case, however, would be that of 
a chain loaded with a load which might be represented by a 
uniformly thick sheet of metal, where the length 00 I is any 
given quantity whatever. A chain so loaded is called a trans- 
formed catenary, and the catenary itself becomes a particular 
case of the transformed catenary. 

We may deduce its equation as follows : — 



783 



APPLIED MECHANICS. 



Let the chain be represented by ACB, and let it be so 
loaded that the load on CD is repre- 
sented by w times area OCDE, so that 
w = weight per unit of area ; then we 
shall have, for this load, 



D 
c ^f 



o E 

Fig. 262. 



= w I ydx 



Hence, from what we have already seen, 



dy_ 
dx 



p w r* _ 

= H=n)j dX 



d 2 y w 

dx 2 = ~H y 



dy d 2 y w dy , 

dx dx 2 H ax 



'rice, integrating, we have 



©■- 



w 
H 



f +c 



dy 



\ But, when — = o,y = a; 



'•-%" 



■• (l)'=5o-->- 



Or, if we write, for brevity, — = m 2 , we have 



w 



ldy\ y 2 — a 2 
\dx) *~ m 2 



dy 



.' — = — \ly 2 — a 2 
dx m 



dy 



dx 



Vy 2 — a 2 m 



.\ log(y + fy - a 2 ) = - + c. 



LINEAR ARCH. 789 



But, when x = o,y s a; 
/. log (a) = c 



bg jO L +.V^)J_ 



*8 



' . ss e*» /. ^ — a 2 = «V W — 2oy*K + J* 



2V jr * & , *. -— \ 

/. -^ = <?/« + <T« /. y = -(*«* + * w ), 

« 2 v 

which is the equation of the transformed catenary. This 
becomes the catenary itself whenever a = m. 

Example. — Given a chain loaded so that the load on CD is 
proportional to the area OEDC. Let OC — 5 feet, BF — 8 feet, 
OF = 4 feet ; weight per unit of area = 80 lbs. Find the 
equation of the transformed catenary, also the tension at C and 
that at B. 

§ 252. Linear Arch. — In all the preceding cases, the chain 
or cord is called upon to resist a tensile stress arising from a 
load that is hung upon it. If, now, the cord be inverted, we 
have the proper equilibrium curve for a load placed upon it, dis- 
tributed in the same manner as before ; only in this latter case 
the cord would be subjected to direct compression throughout 
its whole extent. The equilibrium curve is, then, sometimes 
called a linear arch. The general equation of the equilibrium 
curve remains just as before, 

d 2 y w 
dx 2 = H' 

the axes being so chosen that OX is horizontal and O Y verti- 
cal. 

Thus, if it were required to find the form of the equilibrium 
curve or linear arch, with the upper boundary of the loading 
horizontal, we should obtain a transformed catenary. 



79° APPLIED MECHANICS. 

§ 253. Arches. — In the case of arches composed of a series 
of blocks, as in stone or brick arches, the mathematical treat- 
ment generally used for determining the proper form and 
proportions of the arch has been quite different from that used 
for the determination of the proper form and proportions of 
the iron arch, whether made in one piece, or two pieces hinged 
together, or of a lattice. 

In the case of the iron arch, the treatment involves neces- 
sarily a determination of the stresses acting in all its parts, and 
an adaptation of its form and dimensions to the load, so that at 
no point shall the stress exceed the working-strength of the 
material. 

In the case of the stone arch, it is still a question under 
discussion whether it would not be best to adopt the same 
method, although it would lead to a great deal of complexity, on 
account of the joints. 

Nevertheless, the question usually raised is one merely of 
stability ; i.e., as to the proper form and dimensions to pre- 
vent, not the crushing of the stone, though this must also be 
taken into account if there is any danger of exceeding it, but 
more especially the overturning about some of the joints. 

The question of the stability of the stone arch may present 
itself in either of the two following ways : — 

i°. Given the arch and its load, to determine whether it is 
stable or not. 

2°. Given the distribution of the load, to determine the 
suitable equilibrium curve, and hence the form of arch, suited 
to bear the given load with the greatest economy of material. 

§ 254. Modes of giving Way of Stone Arches. — An arch 
may yield, (i°) by the crushing of the stone, (2 ) by sliding of 
the joints, (3 ) by overturning around a joint. The following 
figures show the modes of giving way of an arch by the last 
two methods. The first two show the dislocation of the arch 
by the slipping of the voussoirs. In the former case the 



FRICTION. 



791 



fiaimches of the arch slide out, and the crown slips down ; in 
the other case the reverse happens. The second two figures 
show the two methods by which an arch may give way by 
rotation of the voussoirs around the joints. 





Fig. 263. 



Fig. 264. 





Fig. 265. 



Fig. 266. 




Before proceeding farther with the problem of the arch, two 
or three matters of a more general nature will be treated, 
which will be necessary in its discussion. 

§ 255. Friction. — Let AB be a plane inclined to the hori- 
zon at an angle 0. Let D be a body resting 
on the plane, of weight DG = W. Resolve 
W into two components, DE and DF respec- 
tively, perpendicular and parallel to the 
plane. The component DE = Wo,o% 6 is 
entirely neutralized by the re-action of the plane ; while DF 
= Wsin 6, on the other hand, is the only force tending to make 
the body slide down the plane. It is an experimental fact, that 
when the angle 6 is less than a certain angle </>, called the 
angle of repose, the body does not slide ; when 6 = <£, the body 
is just on the point of sliding ; and when 6 is greater than <f>, 
the body slides down the plane with an accelerated motion, 
showing that in this case an unbalanced force is acting. This 



Fig. 267. 




9 



79 2 APPLIED MECHANICS. 

angle <f> depends upon the nature of the material of the plane 
and of the body, and on the nature 'of the surfaces. Hence, 
in the first and second cases, the friction actually developed 
by the normal pressure DE just balances the tangential com- 
ponent DF; whereas, in the third case, when the angle of 
inclination of the plane to the horizon is greater than <f>, the 
tangential component DF is only partially balanced by the 
friction. 

Let ab be the plane when inclined to the horizon at an 
d angle <£. The body is then just on the 

point of sliding, hence the component 
df = Wsm<j> is just equal to the fric- 
tion developed between the two surfaces. 
fig. 268. Moreover, if we represent by N the 

normal pressure de = J^cos <f> on the plane, we shall have 

df = JVtan <f>. 

Now, it is an experimental fact, that the friction developed 
between two given surfaces depends only on the normal press- 
ure, i.e., that the friction bears a constant ratio to the normal 
pressure; and since, in this case,- the friction just balances the 
tangential component df = iVtan <j>, the friction due to the 
normal pressure N is 

i^tan <f>. 

Now, it makes no difference what be the position of the 
plane surface : if a normal pressure N be exerted, the friction 
that is capable of being exerted to resist any force F tangential 
to the plane, tending to make the bodies slide upon each other, 
is N tan <f> ; and if the force F is greater than iVtan <j>, the bodies 
will slide, but if F is less than N tan <£, they will not slide. The 
quantity tan <£ is called the co-efficient of friction, and will be 
denoted by /. 




STABILITY OF BUTTRESS ABOUT A PLANE JOINT. ?g$ 

From the preceding it is evident, that, if the resultant press* 
ure on the body makes with the normal to the plane an angle 
less than the angle of repose, the sliding will not take place ; 
whereas, if the resultant force makes with the normal to that 
plane an angle greater than the angle of repose, the body will 
slide. 

§ 256. Stability of Position. — To determine under what 
conditions the stability of the block 
DGHFis secure against turning around 
the edge D: if the resultant of the 
weight of the block and the pressure 
thereon pass outside the edge D, as OR lt 
then the block will overturn ; the mo- 
ment of the couple tending to overturn it i Rz 
being OR, X DE. If, on the other FlG - 26 9- 
hand, it pass within the edge, as OR 2 , the block will not over- 
turn, since the force has a tendency to turn it the opposite 
way around D. Hence, in order that a block may not overturn 
around an edge at a plane joint, the resultant pressure must 
cut the joint within the joint itself. 

In any structure composed of blocks united at plane joints, 
we must have both stability of position and stability of friction 
at each joint, in order that the structure may not give way. 

§257. Stability of a Buttress about a Plane Joint. — 
Let DCEF be a vertical section of a buttress, against which 
a strut rib or piece of framework abuts, exerting a thrust 
P = ZX =z OR. In order that the buttress may not give 
way, it must fulfil the conditions of stability at each joint. Let 
AB be a joint. Should several pressures act against the but- 
tress, the force P in the line ZO may be taken to represent 
the resultant of all the thrusts which act on the buttress above 
the joint AB. Let G be the centre of gravity of the part 
ABEF t and let W = OL be the weight of that part of the 
buttress. Let O be the point of intersection of the line of 



7Q4 



APPLIED MECHANICS. 




Fig. 270. 



direction of the thrust, and of the weight W. Draw the paral- 
lelogram ORNL. Then will ON be the resultant pressure on 
the joint AB : and the conditions of stability require that the 
resultant pressure should cut the joint 
AB at some point between A and B, and 
that its line of direction should make 
with the normal to AB an angle less 
than the angle of repose, <f>; and, in 
order that the buttress may not give way, 
these conditions must be fulfilled at each 
and every joint. 

Another way of expressing this con- 
dition is as follows : The force tending 
to overturn the upper part of the but- 
tress around A is the force F = OR ; 
and its moment around A is F{Ap) = Fp 
if we let Ap = p, whereas the moment of the weight which 
resists this is W(AS) = Wq if we let AS = q. Now, when 
ON passes through A, we have Fp = Wq ; when ON passes 
inside of A, we have Wq > Fp ; when ON passes outside of A, 
we have Wq < Fp. Hence the conditions of stability require 
that 

Wq^Fp or Fp<*Wq. 



Example. — Given a rectangular buttress 
8 feet high, 1 foot wide, and 4 feet thick ; the 
weight of the material being 100 lbs. per 
cubic foot, the buttress being composed of 8 
rectangular blocks 1 X 4 X 1 foot. On this 
buttress is a load of 500 lbs., whose weight' 
acts through K, where OK == 3 feet. Find 
the greatest horizontal pressure P that can be applied along 
the line OK, consistent with stability, against overturning 
around each of the edges a, b t c, d, e, f, g> h. 




Fig. 271. 



LINE GF RESISTANCE IN A STONE ARCH. ?g$ 



Solution. — The weight of each block will be 400 lbs. 
Hence we shall have the following equations : — 



1500 4- 400 x 2 
Stability about a, max P = ■ = 2300. 



b, « = -^ — ■ = i S5 o. 

1500 -f 1200 x 2 
c, « = — = 1300. 

1^00 4- 1600 x 2 

a, « --§ = 1175. 

1500 4- 2000 X 2 

e t " = = 1 100. 

5 

1500 4- 2400 X 2 
" /, " = 1 = io5°- 

1500 4- 2800 X 2 
g, " = ■ " = 1014. 

« h, « = *5oo + 3200 X 2 = ^ 



<c 



« « 



a cc 



It « 



The least of these being 987 lbs., it follows that the great- 
est pressure consistent with stability against overturning is 
987 lbs. 

§ 258. Line of Resistance in a Stone Arch. — In order 
•to solve any problem involving the stability of a stone arch, it 
is necessary that the student should be able to draw a line of 
resistance. To make plain the meaning of the term, the follow- 
ing solution of an example is given. The method of drawing 
the line of resistance employed in this solution is given purely 
for purposes of illustration, and is not recommended for use in 
practice, as a suitable method will be given later. 



; 9 6 



APPLIED MECHANICS. 



EXAMPLE. — Given three blocks of stone of the form shown 

in the figure (Fig. 272), their 
common thickness (perpendic- 
ular to the plane of the paper) 
being such that the weight 
per square inch of area (in the 
plane of the paper) is just one 
pound. 

Given AC = 13 inches, 
BC = 8 inches. Suppose 
these three blocks to be kept 
from overturning by a hori- 
zontal force applied at the 
middle of DE. Find the least, 
value of this horizontal force consistent with stability about 
the inner joints, also its greatest value consistent with stabil- 
ity about the outer joints. 
Solution. 

BK = 16 sin 15° 
AH — 26 sin 15* 
2 I d3) 3 
3((i3) 2 -W : 




Fig. 272. 



CR 



= 4.14112, 
= 6.72932 
" (8) 3 ) _ 



10. 7. 



Altitude of each trapezoid — 5 cos 15 1 
Area of each trapezoid 
Weight of each stone 



^ sin 30 



4.8296. 
26.25 sq. in. 
26.25 ms « 



GG 2 — 8 sin3o° — 10.7 cos i5°sin 15 = 1.325. 

KK 2 = 8 cos 30 — 10.7 cos 15 sin 15 = 4- 2 53- 

KK Z = 10.7 cos 15 cos 45 — 8 cos 30 = 0.380. 

BN 2 = 8 — 10.7 cos i5°sin 15 ■• = 5-3 2 5- 

JBN 3 — 8 — 10.7 cos 15 cos 45 = 0.692. 

BN, = 10.7 cos 2 15° - 8 = 1.983. 

HH 2 — 13 cos 30 — 10.7 cos i5°sin 15 = 8.583. 

HH^ = 13 cos 30 — T0.7 cos i5°cos45° = 3-95°- 



LINE OF RESISTANCE IN A STONE ARCH. 797 

AN 2 = 13 — 10.7 cos 15 sin 15 = 10.325. 

AN Z =13— 10.7 cos 15 cos 45 = 5.692. 

AN 4 = 13 — 10.7 cos 2 15 ' = 3.017. 

G X M = 10.5 — 8COS30 = 3.572. 

K Z M = 10.5 — 8sin30° = 6.500. 

CM = 10.5 = 10.500. 

H X M '= 10.5 — I3sin30° = 4.000. 

Let us represent the thrust at M by 7! Then, to find what 
is the thrust required to produce equilibrium about G, we take 
moments about G, and likewise for the other joints. We may 
proceed as follows : — 

INNER JOINTS. 

Stability about G, 
nG.M) = (26.25) (GG 2 ) 



or 



or 



7(3.572) = (26.25) (1.325) .-. T = 9.74. 

Stability about K, 

T{K,M) = (26.25) {KK 2 - KK Z ) 

7(6. 5 oo) = (26.25) (4.253 - 0.380) • ,\ T= 15.64. 

Stability about B, 

T(CM) = (26.25) (BN 2 + BJV 3 -BN A ) /. T = 10.08 

OUTER JOINTS. 

Stability about H, 

T{H,M) = (z6.2 S )(HH 2 + HH 3 ) ,\ T= 82.25. 

Stability about A, 

T(CM) = (26.25) (AN 2 + AN, + AN 4 ) /. 7=47-59 

It is plain, therefore, that, in order to have equilibrium, the 



79 8 



APPLIED MECHANICS. 



thrust at M must be between 15.64 lbs. and 47.59 lbs. : for, 
if it is less than 15.64 lbs., the arch will turn about an inner 

joint; and if it is greater than 




Pig. 273. 



47.59 lbs., it will turn around 
an outer joint. 

If, now, we draw through 
M a horizontal line to meet 
the vertical drawn through the 
centre of gravity of the first 
stone, and lay off a/3 = 15.64, 
and ay ■— 26.25, then w iH the 
resultant of this thrust a/3 and 
the weight of the first stone ay 
be aS; this being the resultant 
pressure on the joint FG, its 
point of application being «. 
Next, prolong this line aS to 
meet the vertical through the 
centre of gravity of the second 
stone, and combine aS with the 
weight of the second stone, 
thus obtaining, as resultant 
pressure on the joint KH, the 
force £17, whose point of appli- 
cation is at K. Compounding, 
now, £77 with the weight of the 
third stone, we obtain, as final 
resultant pressure on AB, the 
force X/x applied at p. Now, 



coining MzKp by a broken line, we have the Line of Resistance 
corresponding to the thrust 15.64, or the minimum horizontal 
thrust at M. If, now, we construct a line of resistance with 
47.59 lb s -> we obtain the line Ma<f>A, corresponding to maximum 
horizontal thrust at M. 



SYMMETRICAL DISTRIBUTION OF THE LOAD. 799 

If the arch is in equilibrium, and if the horizontal thrust is 
applied at M, it is plain that the actual thrust would either be 
one of these two or else somewhere between these two, and 
hence, that, if the requisite thrust is furnished at M to keep 
the arch in equilibrium, the true line of resistance cannot lie 
outside of these two ; viz., the line corresponding to maximum 
and that corresponding to minimum horizontal thrust at M. 

If the separate stones supported loads, it would be neces- 
sary to take into account these loads, in addition to the weights 
of the stones, in determining the horizontal thrust, and drawing 
the lines of resistance. 

§ 259. Arches with Symmetrical Distribution of the 
Load. — Before considering the conditions of 
stability of an arch, we shall proceed to some 
propositions about lines of resistance corre- 
sponding to maximum and minimum horizon- 
tal thrust. If, in an arch, we draw a line of 
resistance AB through the point A of the 
crown, and then, by changing the horizontal ^|b' 
thrust, we change the line of resistance con- FlG ' 274 ' 

tinuously till it touches the extrados of the arch at C, we 
shall evidently have, in the line AC f B\ a line of resistance 
which has the greatest horizontal thrust of any line that passes 
through Ay and lies wholly within the arch-ring. If, on the 
other hand, we decrease gradually the horizontal thrust until 
the line touches the intrados at D\ then we have in this line 
the line of minimum horizontal thrust that passes through A. 
By lowering the point A, however, and keeping the point C the 
same, we should obtain new lines of resistance with greater 
and greater horizontal thrust ; the greatest being attained when 
the line comes to have one point in common with the intrados. 
Hence a line of maximum horizontal thrust will have one point 
in common with the extrados and one point in common with 
the intrados, the latter being above the former. 




8oo 



APPLIED MECHANICS. 



On the other hand, by retaining the point D' the same, and 
raising the point A, we should decrease the horizontal thrust, 
and thus obtain lines of resistance with less and less horizontal 
thrust ; the least being attained when the line of resistance 
comes to have a point in common with the extrados. Hence 
the minimum line of resistance has a point in common with the 
extrados and one in common with the intrados, the latter being 
below the former. 

These cases are exhibited in the following figures : — 




minimum 
maximum 




minimum, 
maximum 



Fie 275. 



Fig. 276. 



§ 260. Conditions of Stability. — The question of the sta- 
bility of an arch must depend upon the position of its true 
line of resistance. If this true line of resistance lies within 
the arch-ring, the arch will be stable provided the material 
of which it is made is incompressible. If this is not the case, 
the stability of the arch will depend upon how near the true 
line of resistance approaches the edge of the joints ; for the 
nearer it approaches the edge of a joint, the greater the inten- 
sity of the compressive stress at that joint, and the greater the 
danger that the crushing-strength of the stone will be exceeded 
at that joint. Thus, if the true line of resistance cuts any 
given joint at its centre of gravity, the stress upon that joint 
will be uniformly distributed over the joint'. If, however, it 
cuts the joint to one side of its centre of gravity, the intensity 
of the stress will be greater on that side than on the opposite 
side ; and, if it is carried far enough to one side, we may even 
have tension on the other side. 



CRITERION OF SAFETY FOR AN ARCH. 



80 1 



§ 261. Criterion of Safety for an Arch. —There are two 
criteria of safety for an arch, that have been used : — 

i°. That the line of resistance should cut each joint within 
such limits that the crushing-strength of the stone should not 
be exceeded by the stress on any part of the joint. 

2°. That, inasmuch as the joint is not suited to bear tension 
at any point, there should be no tension to resist. 

The distribution of the stress is assumed to be uniformly 
varying from some line in the plane of the joint. The three 
following figures will, on this supposition, represent the three 
cases : — 

i°. When the stress is wholly compression. 

2°. When the stress becomes zero at the edge B. 

3 . When the stress becomes negative or tensile at B* 



*R, 






Fig. 279. 



In all three figures, AB represents the joint which is as- 
sumed to be rectangular in section, AD represents the intensity 
of the stress at A t and BE that of the stress at B ; while R repre- 
sents the point of application of the resultant stress, RR X rep- 
resenting that resultant. 

Proposition. — If the stress on a rectangular joint vary 
uniformly from a line parallel to one edge, the condition that 
there shall be no tension on any part requires that the result- 
ant of the compressive stress shall be limited to the middle 
third of the joint. 

Proof. — Let AB (Fig. 278) represent the projection of 
the ioint on the plane of the paper. It is assumed that the 



802 APPLIED MECHANICS. 

stress is uniformly varying ; and, if there is to be no tension 
anywhere, the intensity at one edge must not have a value less 
than zero, hence at the limiting case the value must be zero ; 
hence this limiting case is correctly represented by the figure, 
and the resultant of the compression will be for this case at the 
centre of stress. Thus, if AD represent the greatest intensity 
of the stress, then we shall have, if B be the origin and BA the 
axis of x, if the axis of y be perpendicular to AB at B, and if 
we let a = intensity of stress at a unit's distance from B, that 
RR X = affxdxdy, and (BR) (RR T ) = affx 2 dxdy; 

f fx 2 dxdy 2 

' SJxdxdy ~ . M" ~ ¥ *' 

2 

if b = breadth, and h = BA = height of rectangle. 

Hence, if the resultant of the compression be nearer A than 
R, there will be tension at B ; and, on the other hand, if it be 
nearer B than J/z, there will be tension at A. Hence follows 
the proposition as already stated. 

While the above is probably the condition most generally 
used to determine the stability of an arch, at the same time, if 
there is any danger that the intensity of the stress at any part 
of any joint may exceed the working compressive strength of 
the stone, this ought to be examined, and hence a formula by 
which it may be done will be deduced. 

Let AB (Fig. 279) be the joint, and let, as before, b be its 
breadth, and h == AB = depth ; then, suppose the pressure to 
be uniformly varying, DA == f == the working-strength per unit 
of area == greatest allowable intensity of compression ; then the 
entire stress on the joint will be represented by the triangle 
A CD, for the joint is incapable of resisting tension. 
Hence 

AR = \AC /. AC = 2>AR; 



POSITION OF THE TRUE LINE OF RESISTANCE. 803 

but 

and this is the least distance from the outer edge at which the 
resultant should cut the joint. 

We thus obtain, in terms of the pressure on any joint, and 
of the working-strength of the material, the limits within which 
the line of resistance should pass, in order that the working- 
strength of the stone may not be exceeded. 

§ 262. Position of the True Line of Resistance. — The 
question of the most probable position of the true line of 
resistance involves the discussion of the properties of the 
elastic arch. This discussion will be given later ; but, for the 
present, the statement only of the following proposition, due to 
Dr. Winkler, will be given : — 

"For an arch of constant section, that line of resistance is 
approximately the true one zvhich lies nearest to the axis of the 
arch-ring, as determined by the method of least squares." 

From this it will follow : — 

i°. That, if a line of resistance can be drawn in the arch- 
ring, then the true line of resistance will lie in the arch-ring ; 
and 

2°. That, if a line of resistance can be drawn within the 
middle third of the arch-ring, then the true line of resistance 
will lie in the middle third. 

But, before proving this proposition, the proposition will be 
used, and the method explained, for determining whether a line 
of resistance can be drawn within the arch-ring : for, if it can, 
then the true line of resistance must lie within the arch-ring ; 
and if no line of resistance can be drawn within the arch-ring, 
then the true line of resistance cannot pass within the arch- 
ring, and the arch would necessarily be unstable, even if the 
materials were incompressible. 

By following the same method, we could determine whether 



804 APPLIED MECHANICS. 

it was possible to draw a line of resistance within the middle 
third of the arch-ring ; and, if this is found to be possible, we 
should know that the true line of resistance will pass within 
the middle third of the arch-ring. 

Hence our most usual criterion of the stability of a stone 
arch is, whether a line of resistance can be passed within the 
middle third of the arch-ring. 

If the condition be used, that the working-strength of the 
stone for compression be not exceeded, then, instead of the 
middle third, we shall have some other limits. 

In what follows, an explanation will be given of Dr. Scheff- 
ler's method (that most commonly employed) of determining 
whether a line of resistance can be drawn within the arch-ring, 
inasmuch as the same method can be employed to determine 
whether such a line can be drawn within the middle third or 
within any other given limits. 

§ 263. Preliminary Proposition referring to Arches Sym- 
metrical in Form and Loading. — An arch audits load being 
given, a line of resistance can always be made to pass through 
any two given points ; hence, if any two points of a line of 
resistance are given, the line is determined. 

Proof. — Let the arch be that shown in Fig. 281 ; and let us 
consider first the special case when the two given points are A, 
the top of the crown-joint, and G 4 , the foot of the springing- 
joint. In this case, the only quantity to be determined is the 
thrust at A. Let this thrust be denoted by T ; let P be the 
total weight of the half-arch and its load ; let a be the perpen- 
dicular distance of the point G 4 from a vertical line through the 
centre of gravity of the entire half-arch and its load ; let h be 
the vertical depth of G 4 below A. Then, taking moments about 
G 4 , we must have 



Th = Pa 
„ Pa 



H> « 



DR. SCHEFFLER'S METHOD. 805 

and the line of resistance can then be drawn with this thrust, 
as has been done in the figure. Next take the general case, 
when the given points are not in these special positions. Let 
them be any two points, as A 2 and G 3 . 

In this case, the point of application of the thrust at the 
crown is not necessarily A, but may be some other point of the 
crown-joint : hence the quantities to be determined are two ; 
viz., the thrust T at the crown, and the distance x of its point 
of application below A. Let the combined weight of the first 
two voussoirs and their load be P t , and the horizontal distance 
of A 2 from a vertical line through the centre of gravity of P x be 
a s . 

Let P 2 be the combined weight of the first three voussoirs 
and their load, and let a z be the horizontal distance of G 3 from 
a vertical line through the centre of gravity of P 2 . 

Let the vertical depth of A 2 below A be k lt and that of G 3 
below A be k 2 . Then, taking moments about A 2 and G 3 respec- 
tively, we shall have 

T{h x — x) = P,a, and T(k 2 - x) = P 2 a 2 , 

two equations to determine the two unknown quantities T and 
x, which can easily be solved in any special case ; and'the result- 
ing line of resistance can be drawn, which will pass through the 
two given points. 

§ 264. Scheffler's Method. — In using Scheffler's method 
of determining whether it is possible to pass a line of resistance 
within a given portion of the arch-ring as the middle third or 
not, we should proceed as follows ; viz., — 

First pass a line of resistance through 1, the top of the 
middle third of the crown-joint (Fig. 280), and <?, the inside of 
the middle third of the springing-joint. If this line lies wholly 
within the middle third, it proves that a line of resistance can 
be drawn within the middle third. 

If this line of resistance does not pass entirely within the 



806 APPLIED MECHANICS. 

middle third, proceed as follows : Suppose the line thus drawn 

to be \abcde, passing without the middle third on both sides, 

a /jj as shown in the figure. Then from #, the point 

a /?Z/T where it is farthest from the extrados of the 

/w/ i middle third, draw a normal to this extrados, and 

f^U find the point where this normal cuts this extra- 

, UMd ! dos : in this case, 2 # is the point in question. In 

| ( Ifl this way determine also the point 7, where the 

normal from d cuts the intrados of the middle 

Fig. 280. i«ii 1 • r 1 1 

third ; then pass a new line of resistance through 
the points 2 and 7, .determining the thrust and its point of ap- 
plication. If this new line of resistance lies within the middle 
third, then it is plain that it is possible to draw a line of resist- 
ance within the middle third ; if not, it is not at all probable 
that it is possible to draw such a line. 

If the line of resistance drawn through 1 and e goes outside 
the middle third only beyond its extrados, as at a, we should 
draw our second line of resistance through 2 and e ; if, on the 
other hand, it goes outside only below the intrados of the 
middle third, as at d t we should draw our second line through 
1 and 7. 

In the construction, we make use of a slice of the arch in- 
cluded between two vertical planes a unit of distance apart ; and 
we take for our unit of weight the weight of one cubic unit of 
the material of the voussoirs, so that the number of units of 
area in any portion of the face of the arch shall represent the 
weight of that portion of the arch. 

We next draw, above the arch, a line (DD 4 in Fig. 281), 
straight or curved, such that the area included between any 
portion of it, as D^D„ the two verticals at the ends of that por- 
tion, and the extrados of the arch-ring, shall represent by its 
area the load upon the portion of the arch immediately below 
it. This line will limit the load itself whenever this is of the 
same material as the voussoirs ; otherwise it will not. We shall 
always call it, however, the extrados of the load. 



DR. SCHEFFLER'S METHOD. 



807 



The mode of procedure will best be made plain by the solu- 
tion of examples ; and two will be taken, in the first of which 
only one trial is necessary to construct a line of resistance that 
shall lie wholly within the arch-ring, and, in the second, two 
trials are necessary. 

Example. — The half-arch under consideration is shown in 
Fig. 281, GG 4 being the intrados, AA 4 the extrados of the arch, 




Fig. 281. 



ind DD^ the extrados of the load. The arcs GG 4 and AA 4 are 
concentric circular arcs. The data are as follows : — 

Span = 2(G 4 0) = 6.00 feet, 

Rise = GO =0.50 foot, 

Thickness of voussoirs = AG = A 4 G 4 =0.75 foot, 

Height of extrados of load above A — AD — 0.80 foot. 



The position of the joints is not assumed to be located. We 
therefore draw through A a horizontal line AB, and divide this 
into lengths nearly equal, unless, as is usual near the springing, 
there is special reason to the contrary. Thus, we make the first 
three lengths each equal to 1 foot, and thus reach a vertical 



8o8 



APPLIED MECHANICS. 



through G 4 ; and then the last division has a length of 0.24 foot. 
We have thus divided the half-arch and its load into four parts ; 
viz., GDD X H„ H X D^H M H 2 D 2 D 2 G A , and G A D Z D A A„ the loads 
on these respective portions being represented by their areas 
respectively. We assume the centre of gravity of each load to 
lie on its middle vertical ; and we then proceed to determine the 
numerical values of the several loads, the distances of their 
centres of gravity from a vertical through the crown, also the 
amount and centre of gravity of the first and second loads 
together, then of the first, second, and third, etc. 

The work for this purpose is arranged as follows : — 



(1) 


(2) 


(3) 


(4) 


(5) 


(6) 


(7) 


(8) 


(9) 


(10) 




Length. 


Height. 


Area. 


Lever 
Arm. 


Moment. 


Partial Sums. 


Area. 


Moment. 


Lever 

Arm. 


1 

2 
3 
4 


1. 00 
I. OO 
1. 00 
O.24 


1-57 

1.68 
1.90 

1.72 


I.570 
I.680 
I.900 
O.413 


O.50 
I.50 
2.50 
3.12 


O.785 
2.520 

4-75°' 
1.287 


I 

1 + 2 

1 + 2+3 
1+2+3+4 


i-57o 
3.250 
5.150 
5-563 


O.785 
3-305 
8.055 
9-342 


O.500 
I. OI 7 

1.563 
I.680 


- 


3- 2 4 


- 


S-563 


- 


9-342 


- 


- 


- 


- 



Column (1) shows the number of the voussoir. 

" (2) gives the horizontal lengths of the several trape- 
zoids. 

" (3) gives the middle heights of the trapezoids. 

" (4) gives the areas of the trapezoids, and is obtained by 
multiplying together the numbers in (2) and (3). 

" (5) gives the distances from A to the middle lines of 
the trapezoids. 

" (6) gives the products of (4) and (5), giving the moments 
of the respective loads about an axis through A 
perpendicular to the plane of the paper. 



DR. SCHEFFLER'S METHOD. 



809 



Column (7) merely indicates the successive combinations of 
voussoirs. 
" (8) has for its number?, — 

i°. The area representing the first load. 
2°. The area representing the first two loads. 
3°. The area representing the first three. 
4 . The area representing the first four. 
" (9) has for its numbers, — 

i°. The moment of the first load about A. 
2°. The moment of the first and second loads 

about A. 
3°. The moment of the first, second, and third 

loads about A. 
4 . The moment of the first, second, third, and 
fourth loads about A. 
(10) is obtained by dividing column (9) by column (8) ; 
the quotients being respectively the distance 
from A to the centres of gravity of the first, of 
the first and second, of the first, second, and 
third, and of the first, second, third, and fourth 
loads. 
The calculation thus far is purely mathematical, and merely 
furnishes us with the loads and their points of application ; in 
other words, furnishes us the data with which to begin our 
calculation of the thrust. Before passing to this, it should be 
said, however, that we now assume the joints to be drawn 
through the points A 1} A 2 , A v and A 4 , and generally normal to 
the extrados of the arch. 

In this proceeding, we, of course, make an error which is 
very small near the crown and increases near the springing of 
the arch ; this error, in the case of voussoir A Y A 2 G X G 2 , amounts 
to the difference of the two triangles A 2 G 2 H 2 and A^G^H^ A 
manner of making a correction by moving the joint will be 
explained later ; but now we will complete our example, as the 



8 1 A P PLIED MECHA NICS. 

errors are not serious in this example. We now pass a line of 
resistance through a, the upper point of the middle third of 
the crown-joint, and /? 4 , the lowest point of the middle third of 
the springing. For this purpose take moments about /3 i ; and 
we shall have, if T =. thrust at the crown, 

0-75^ = (5-563) (3.075 ~ 1-68). 

since 5.563 is the whole weight, and 3.075 — 1.68 is its leverage 
about /? 4 . 
Hence 

0.75^= (5.563) (1.395) = 7760. 
.-. T= 10.35. 

Hence we proceed to draw a line of resistance through a, 
assuming, as the horizontal thrust, 10.35. To do this we pro- 
ceed as follows: From #, the point of intersection of ay 
with the vertical through the centre of gravity of the first 
trapezoid, we lay off ab to scale equal to 10.35, an ^ then lay 
off bC vertically to scale equal to 1.57, the first load ; then will 
Ca be the resultant pressure on joint A x G iy and its point of 
application will be P, which gives us one point in the line of 
resistance. To obtain the point P Jf we lay off aa t = 1.017, 
the lever arm of the first two loads ; then lay off a x b x = 10.35, 
the thrust; then lay off b l C l equal to 3.25, the weight of the 
first two loads. Then will C 1 a l be the pressure on the second 
joint ; and the point P x , or its point of application, is at the 
intersection of C x a x with A^G^. 

Then lay off aa 2 — 1.563, ajb^ — 10.35, £ 3 C 2 == 5. 150; and 
P Q , the next point of the line of resistance, is the intersection 
of £7 2 tf 2 with A 3 G Z . Then lay off aa % = 1.680, a z b 3 — 10.35, 
b % C % = 5.563 ; and C z a z is the pressure on the springing, and 
this will intersect A 4 G 4 at /? 4 unless some mistake has been 
made in the work. Then is aPP i P i (3 A the line of resistance 
through a and /? 4 . and this lies entirely within the middle third. 



SCHEFFLER'S MODE OF CORRECTING THE JOINTS. 8ll 



Henee we conclude that it is possible to draw a line of resist- 
ance within the arch-ring without having recourse to another 
trial. 

§ 265. Scheffler's Mode of Correcting the Joints. — The 
following is the approximate construction given 
by Scheffler for correcting the joint : Let DCG 
be the side of the trapezoid, and CH the uncor- 
rected joint. From b, the middle point of GH, 
draw Db; then draw Gc parallel to bD, and ch 
parallel to CH. Then will ch be the corrected 





Fig. 282. 



joint. 

Conversely, having given the 

joint CH, to find the side of the trapezoid which 

limits the portion of the load upon it : through 

C draw DG vertical ; join D with b, the middle 

point of GH; then draw Cg parallel to Db ; 

then, from g, drawing dg vertical, we thus have 

the desired side of the trapezoid. 

§ 266. Another Example. — Another example will now be 

solved, which necessitates two trials, and where some of the 

joints have to be corrected. It is practically one of Schemer's. 

The dimensions of the arch are as follows : — 



Fig. 283. 



Half-span 3 2 -97 feet. 

Rise 24.74 feet. 

Thickness of ring 5.15 feet. 

Height of load at crown 8.24 feet. 

Height of load at springing 33-5° f eet - 



The arch may be drawn by using, for the intrados, two 
circular arcs. Beginning at the springing, draw a 6o° arc with 
a radius of one-fourth the span ; then, with an arc tangent to 
this arc, continue to the crown, the proper rise having been 
previously laid off. The work for drawing a line of resistance 



8l2 



APPLIED MECHANICS. 



through the top of the middle third of crown-joint and the 
inside of the middle third of the springing will be given with- 
out comment. It is as follows : 



(1) 


(2) 


(3) 


(4) 


(5) 


(6) 


(7) 


(8) 


(9) 


(10) 


*S '5 


Length. 


Height. 


Area. 


Lever 


Mo- 


Partial Sums. 


Area. 


Moment. 


Lever 










Arm. 


ment. 








Arm. 


I 


8.24 


I4.I 


1 16.18 


4.12 


479 


! 


1 16.18 


479 


4.12 


2 


8.24 


164 


135-14 


12.36 


1670 


1 + 2 


251.32 


. 2149 


8.55 


3 


8.24 


18.3 


150.79 


20.6o 


3106 


1 + ...+3 


402.II 


5255 


13.07 


4 


4.13 


22.6 


93-34 


26.79 


2500 


i+. . .+4 


49545 


7755 


r 5- 6 5 


5 


4.13 


27.1 


1 1 1.92 


30.92 


3461 


1+...+5 


607.37 


11216 


18.46 


6 


5-14 


34-7 


178.36 


35-55 


6341 


1+...+6 


78573 


J7557 


22.34 


- 


- 


- 


78573 


- 


17557 


- 


- 


- 


- 



28.5 T = (78573) (34-9 - 22.34), 

28.5 T = 9868.77; 

.-. T = 346.27. 

Hence we construct the line of resistance passing through the 
top of the middle third of crown-joint and the inside of the 
middle third of the springing, using the thrust 346.27. 

The construction is shown in the figure, and is entirely 
similar to that previously used. The student will readily 
identify this first line of resistance, and will see that it goes 
outside the middle third both above and below, being farthest 
above the extrados at the first joint from the crown, and farthest 
inside of the intrados opposite the first joint from the spring- 
ing. Hence we proceed to pass a new line of resistance through 
the top of the middle third of the first joint from the crown, 
and the inside of the middle third of the first joint from the 
springing. 



SCHEFFLER'S MODE OF CORRECTING THE JOINTS. 813 

For this purpose we do not need to make out a new table, 
as it is not necessary to insert any new joints. We need only 
two more dimensions, i.e., the vertical depth of each of these 




Fig. 284. 

points below the top of the crown : these depths are respec- 
tively 2.10 and 16.8. 

Hence we proceed as follows : 
Let T =f thrust at the crown ; 

x = distance of its point of application below the top of 
the crown-joint. 



8 14 APPLIED MECHANICS. 

i°. Take moments about the upper one of the two points, 
and we have 

7(2.10 — x) = (116.18) (8 — 4.12) = 450.778. 

2°. Take moments about the lower one of the two points, 
and we have 

r(i6.8 - x) = (607.37) (30.5 - 18.46) = 7312.734. 

Solving these equations, we obtain 

T = 466.8, x = 1. 1 34. 
Hence through a point on the crown-joint at a distance 1.134 
below the top of the middle third of the crown-joint draw 
a horizontal line, this line being the line of action of the thrust. 
Then, making the construction for a new line of resistance just 
as before, only using this new point of application of the thrust, 
and using for thrust 466.8, we shall obtain a new line of resist- 
ance, which passes through the desired points. 

Another method of drawing a line of resistance frequently 
pursued is the following : — 

After determining the loads on the successive voussoirs, 
and also the thrust for the particular line of resistance which 
we wish to draw, layoff these loads and thrust to scale in their 
proper order and directions, and construct a force polygon 
{see § 126), then construct the corresponding frame (equilib- 
rium polygon), which shall have its apices on the vertical lines 
passing through the centres of gravity of the loads as drawn in 
the figure of the arch. The intersections of the lines of the 
frame with the joints give us points of the line of resistance, 
and the line of resistance can be drawn by joining them. 

Thus, applying the solution to Fig. 281, we lay off 
ab = 1.570, m • be = 1.680, 

cd = 1.900, de = 0.413. 

Also, lay off 

oa = 5.87, 

and draw the lines ob, oc, od, and oe. 



DRA WING A LINE OF RESISTANCE. 



815 



T 




1 
1 
1 
| 


1 
1 
1 
1 




1 
1 
1 


4- 


\P^ 


1 


1 

1 

J&— H — 




|0 A 


EI 


1 


t 1 


g\- 








1 
1 

— — — i 


^a\ 













8i6 



APPLIED MECHANICS. 



Then from A draw Aa parallel to oa, then draw aa x parallel 
to ob t a^a^ parallel to oc, a 2 a 3 parallel to od, and a 3 a A parallel to 
oe, this last prolonged backwards of course* passes through G A ; 
then will the line of resistance be obtained by joining the points 
AP X P 2 P % G . 

The last figure on preceding page shows the same method 
applied to Fig. 284. 

§ 267. Examples. — Four more examples will now be given 
to be worked out by the student. The dimensions are approx- 
imately those given in some of Scheffler's examples. 

Example I. — Half-span = CD = 65.16 feet, rise = FD = 
13.85 feet, AF ' = 5.32 feet, AE = 6.40 feet. The arcs CF and 
AG are concentric circular arcs. Given width of first five 
horizontal divisions of line AB, counting from A, each 10.66 
feet; width of sixth division, 11.86 feet; of seventh, 2.2 feet. 
Determine the possibility of drawing a line of resistance in the 
arch-ring. 




Fig. 285. 



Example II. — Half-span = 63.98 feet, rise = FD = 31.99 
feet, AF = CG = 5.32 feet, AE =2.13 feet. The intrados and 
extrados of this arch are seven centred ovals, both drawn from 



EXAMPLES. 



817 



the same centres. Beginning at tne springing, an arc with a 
radius of 21 feet is drawn, subtending 39 ; the curve is con- 
tinued by a curve subtending 24 , and having a radius of 35.55 
feet. From F an arc subtending io° is drawn from a centre 
on FD produced, and with a radius of 152 feet; the curve is 
completed by an arc connecting the second and last. 




Fig. 286. 



Given horizontal width of each of first six divisions, counting 
from A, 10.66 feet; horizontal width of seventh division, 5.32 
feet. Determine the possibility of drawing a line of resistance 
in the arch-ring. 




Fig. 287. 



Example III. — Given span = 74.18 feet; rise = 45.83 



8i8 



APPLIED MECHANICS. 



feet ; radius of intrados = 82.42 feet ; radius of extrados = 
91.18 feet ; height of load at crown = 8.24 feet ; width of each, 
of five divisions nearest crown = 8.24 feet; width of sixth 
stone = 4.13 feet. Determine the possibility of drawing a line 
of resistance within the arch-ring. 

Example IV. — Given span = 37.07 feet ; thickness of 
ring = AB = 3.08 feet ; height of load = EC = 82.42 feet. 
Determine the possibility of drawing a line of resistance within 
the arch-ringc 




Fig. 288. 



§ 268. Griterion of Stability. — It has already been stated, 
that, if a line of resistance can be drawn within the arch-ring, 
then the true line of resistance will lie within the arch-ring. 



UNSYMMETRICAL ARRANGEMENT. 819 

With those who, like Scheffler, consider the material of the 
voussoirs incompressible, the criterion of stability of an arch is, 
that it should be possible to draw aline of resistance within the 
arch-ring. 

On the other hand, Rankine would decide upon the stability 
of an arch by determining whether a line of resistance can be 
drawn within the middle third of the arch-ring. 

Other limits have been adopted instead of the middle third. 
In some cases the only reason for deciding upon what these 
limits should be has been custom or precedent. 

They might also be determined so that there should be no 
danger of exceeding the crushing-strength of the stone. 

It is needless to say that the first method is incorrect ; for 
the material of the voussoirs is never incompressible, and an 
arch where the true line of resistance touches the intrados or 
extrados could not stand, as the stone would be crushed. 

Nevertheless, no example will be solved here, where 
we determine the possibility of drawing a line of resist- 
ance within any other limits than the middle third, as the 
method of procedure is entirely similar to what we have done, 
the computation of the entire table being the same in all cases, 
the only difference occurring in the computation of the thrust 
and its point of application, and the consequent construction of 
the line of resistance. The method to be pursued is, as before, 
by taking moments about the points through which it is desired 
that the line of resistance shall pass. 

§ 269. Unsymmetrical Arrangement. — When the arch is 
unsymmetrical, either in form or loading, the same criterion as 
to being able to pass a line of resistance within the middle thirtf 
or other limits of the arch-ring will serve to determine its sta 
bility. The method of procedure differs, however, from the fact, 
that whereas we have heretofore found it necessary to study 
only the half-arch and its load, and have had the advantage of 
knowing, from the symmetry of arch and load, that the thrust at 



820 



APPLIED MECHANICS. 




Fig. 2 



the crown is horizontal, we have not that advantage here, and 
hence we must study the entire arch, and we must assume that 
the thrust at the crown may be oblique, and hence have a verti- 
cal as well as a horizontal component. 

In this case it will be necessary to have three instead of two 
points given, in order to determine a line of resistance. 

If we assume (Fig. 289) a vertical joint at the crown, and let 
P = vertical component of the thrust at 
the crown, A = horizontal component of 
the thrust at the crown, x = distance 
of point of application of thrust at the 
crown below upper point of crown-joint, 
we have thus three unknown quantities, 
and we shall therefore need three equa- 
tions to determine them. 

In this case, therefore, we must have three points of the line 
of resistance given, in order to determine it ; and a reasoning 
similar to that pursued in § 263 would show that a line of re- 
sistance can always be passed through any three given points. 

In performing the work, we should need to make out a table 
for the part of the arch on each side of the crown-joint, show- 
ing the loads, and centres of gravity of the loads, on each vous- 
scir, and on combinations of the first two, first three, etc. ; this 
portion of the work being entirely similar to that done in the 
case of arches of symmetrical form and loading, only that we 
require a separate table for the parts on each side of the crown- 
joint. 

When these two tables have been worked out, we next pro- 
ceed to impose the conditions of equilibrium by taking moments 
about each of the three points given. 

Thus, suppose that (as is usually done first) we pass a line 
of resistance through the top of the middle third of the crown- 
joint and the inside of the middle third of each springing- 
joint, we then have only two unknown quantities to determine ; 



C/JVS YMME 1 'Rl CA L A RRA N GEM EN T. 821 

viz. P and <2, inasmuch as x becomes zero. Hence we take 
moments about the inner edge of the middle third of each of 
the springing-joints. 

In taking moments about the inner edge of the middle 
third of the left-hand springing-joint, we impose the conditions 
of equilibrium upon the forces acting on that part of the arch 
that lies to the left of the crown-joint. These forces are : (i°) 
its load and weight, which tend to cause right-handed rotation ; 
(2°) the horizontal component of the thrust exerted by the 
right-hand portion upon the left-hand portion ; (3 ) the vertical 
component P of the thrust exerted by the right-hand portion 
upon the left-hand portion. 

It is necessary to adopt some convention, in regard to the 
sign of Py to avoid confusion : and it will be called positive 
when the vertical component of the thrust exerted by the right- 
hand portion on the left-hand portion is upwards ; when the 
reverse is the case, it is negative. 

We next take moments about the inner edge of the middle 
third of the right-hand springing-joint, and impose the condi- 
tions of equilibrium upon the forces acting upon the right-hand 
portion of the arch. In doing this, we must observe that we 
have for these forces, (i°) the weight and load which tend to 
cause left-handed rotation ; (2°) the horizontal component Q 
of the thrust exerted by the left-hand portion upon the right- 
hand portion, — this acts towards the right ; (3 ) the vertical 
component P of the thrust exerted by the left-hand portion 
upon the right-hand portion ; and this, when positive, acts 
downwards. 

Having determined the values of Q and P, we next proceed 
to draw the line of resistance, by the use of either of the 
methods employed, with symmetrical arches, observing only 
that the thrust, i.e., the resultant of P and Q, is now oblique, 
and that it acts in opposite directions on the two sides of the 
crown-joint. 



822 



APPLIED MECHANICS. 



Having drawn this line of resistance, if we find that it passes 
outside of the middle third, we draw normals through the points 
where it is farthest from the middle third, and thus obtain 
three points through which to draw a line of resistance : then, 
taking moments about each of these three points, we deter- 
mine, from the three resulting equations, values of Q, P y and 
x> and proceed to draw our new line of resistance; and, if this 
does not pass entirely within the middle third, it is not at all 
probable that a line of resistance can be drawn within the 
middle third. All the above will be made clearer by the fol- 
lowing example : 

Example. — Given an unsymmetrical circular arch, shown 
in the figure, the intrados and extrados being concentric 
circles, EM = 4', HF = i'.85, radius of EHF — 6, AH = o'.5, 
AK == o'.8, to determine the possibility of drawing a line of 




Fig. 290. 



resistance in the arch-ring. The tables following show the 
mode of dividing up the load, and getting the centres of 
gravity, also the mode of arranging the work for this pur- 
pose. 



C/NS YMME TRICA L ARRANGEMENT. 



823 



LEFT-HAND PORTION. 



£•5 

3 O 


Width. 


Height. 


Area. 


Lever 
Arm. 


Moment. 


Partial 
Sums. 


Area. 


Moment. 


Lever 
Arm. 


I 

2 

3 

4 
5 


I. OO 
I. OO 

I. OO 
I. OO 
o-33 


I.32 
I.48 
I.84 
2.42 
2.63 


I.32 
I.48 
I.84 
2.42 
O.87 


O.50 
I.50 
2.50 

3-5° 
4.17 


O.660 
2.220 
4.600 
8.470 
3.628 


I 

I + 2 

1 + 2 + 3 
1 + ... + 4 
1 + ...+ 5 


I.32 
2.8o 
4.64 
7.06 

7-93 


O.660 

2.880 

7.480 

I5-950 

19.578 


O.50 
I.03 
I.6l 
2.26 
2.47 


- 


- 


- 


7-93 


- 


19.578 


- 


- 


- 


- 



RIGHT-HAND PORTION. 



Vh 

2.s" 

S3 


Width. 


Height. 


Area. 


Lever 
Arm. 


Moment. 


Partial 
Sums. 


Area. 


Moment. 


Lever 
Arm. 


I 

2 


1. 00 

1. 00 


I.32 
I.48 


I.32 

I.48 


O.50 
I.50 


O.660 
2.220 


I 

I + 2 


I.32 
2.8o 


O.660 
2.880 


O.50 
I.03 


- 


- 


- 


2.80 


- 


2.88o 


- 


- 


- 


- 



Now take moments about the inner edge of the middle 
third of the left-hand springing, and we have 

1-730 — 4-ioP = 7.93(4-10 - 2.47) = 12.9259. 

Then take moments about the inner edge of the middle third 
of the right-hand springing, and we have 

0.472 — 1.90P = 2.80(1.90 — 1.03) = 2.4360. 

Solving these two equations gives us 

Q = 6.246. 

P = 0.263. 



824 APPLIED MECHANICS. 

If R represents the resultant of P and Q, we have 



R = V P* + Q 1 =6.251; 

hence we proceed, as follows, to pass a line of resistance through 
the top of the middle third of the crown-joint and the inner 
edge of the middle third of each springing : 

Through the top of the middle third draw a horizontal line* 
Lay off aa = 6.626 and ab = 0.263, anc * draw ab ; then ab = 
6.635 represents, in direction and magnitude, the thrust at 
the crown. Using this thrust in the same way as we did the 
horizontal thrust in the case of symmetrical arches, we obtain 
the line of resistance which is farthest outside of the arch at 
d; hence, drawing a normal to the arch from d, we obtain c, the 
upper edge of the middle third of the first joint from the 
crown. Hence we proceed to pass a new line of resistance 
through B y c, and y. 

To do this we must assume Q, P, and x all unknown. 

i°. Take moments about £, and we have 

(1.73 -*)<2 + 4.1^= 12.9259- 

2°. Take moments about y, and we have 

(0.47 -x)Q—i.9P= 2.436. 

3 . Take moments about c, and we have 

(0.078 - x)Q + P = (1.32) (0.45) = 0.594. 

Solving these three equations, we obtain 

Q = 6.905, 

P = 0.297, 

x = 0.035. 

Hence . 

R = V P* + ff = 6.91. 

Hence, if we lay off a distance 0.035 below a, we shall have 
the point on the crown-joint at which the thrust is applied ; 



GENERAL REMARKS. 82$ 

and making the same kind of construction as we just made, 
only using this point instead of A, and these new values of Q 
and P, we construct the second line of resistance. The con- 
struction is omitted in order not to confuse the figure ; but the 
line of resistance is drawn, and the student can easily make 
the construction for himself. It will be seen, that, in this 
case, this new line of resistance lies entirely within the arch- 
ring. 

§ 270. General Remarks. — Whenever there are also hor- 
izontal external forces acting upon the arch, these should be 
taken into account in imposing the conditions of equilibrium. 

It will be noticed, that, in the preceding discussion, it has 
always been assumed that the load upon any one voussoir is the 
weight of the material directly over that voussoir. This is the 
assumption usually made in computing bridge arches : and it 
may be nearly true when the height of the load above the crown 
is not great ; but even then it is not strictly true, and when 
this depth becomes great, as would be the case with an arch 
which supports the wall of a building, it is far from true, as the 
distribution of the load actually coming upon different parts of 
the arch must vary with, and depend upon, the bonding of the 
masonry, and also upon the co-efficient of fricticfn of the mate- 
Hal. Thus, in the case of an arch supporting a part of the wall 
«f a building, it is probable that the only part of the load that 
-:omes upon the arch is a small triangular-shaped piece directly 
)ver the arch, and that above this the material of the wall is 
supported independently of the arch. This will be plain when 
we consider, that, were such an arch removed, the wall would 
remain standing, only a few of the bricks near the arch falling 
down ; and though the number of bricks that would fall would 
be greater while the mortar is green, still even then only a few 
would drop out. 

In regard to these matters, we need experiments ; but thus 
far we have none that are reliable. 



826 APPLIED MECHANICS. 

Then, again, we have arches supporting a mass of sand or 
gravel ; and then the mutual friction of the particles on each 
other comes into play, and it is not true in this case that the 
load on any voussoir is the weight of the material directly 
above that voussoir. In some cases this has been accounted 
as a mass of water pressing normally upon the arch, but we 
cannot assert that such a course is correct. 

On the other hand, there are cases where we know that an 
arch is subjected to horizontal as well as to vertical forces, and 
sometimes we cannot tell how great these horizontal forces are. 
Thus, the forms of sewers are an arch for the top and an 
inverted arch for the bottom ; but in this case the sides of the 
ditch in which the sewer is laid when building it, are capable of 
furnishing whatever horizontal thrust is needed to force the 
line of resistance into the arch-ring, provided that a horizontal 
thrust is what is needed to force it in. Hence it is, that, were 
the attempt made to pass a line of resistance within the arch- 
ring of almost any successful sewer, accounting the load as the 
weight of the earth above it, the line would almost invariably 
go outside ; but the earth on the sides is capable of furnishing 
the necessary horizontal thrust to force it inside, unless a care- 
less workman has omitted to ram it tight, or unless some other 
cause has loosened it on the si'des of the sewer. 

If we know, in any case, the actual law of the distribution 
of the load, we can determine the proper form for the arch by 
the methods of the first part of this chapter, as was done in 
the case of the parabola and of the catenary. Scheffler's 
method is, however, the one almost always used for determin- 
ing the stability of any stone arch against overturning around 
the joints. 

Should there ever arise a case where there was danger that 
the resultant pressure on any joint made an angle with the 
joint greater than the angle of friction, this could be remedied 
by merely changing the inclination of the joint. 



GENERAL THEORY OF THE ELASTIC ARCH 827 

§ 271. General Theory of the Elastic Arch. — In the case 
of the iron arch, the loads upon the arch are all definitely 
known ; and it is necessary to ascertain with certainty the stress 
in all parts of the structure, and to so proportion the different 
members as to bear with safety their respective stresses. 

The general discussion of the method used in calculating 
such arches will now be given ; the method used being practi- 
cally that followed by Dr. Jacob J. Weyrauch, and explained 
more at length in his "Theorie der Elastigen Bogentrager." 

This discussion is also necessary in order to prove the 
proposition already enunciated in § 262 ; viz., that "for an arch 
of constant cross-section, that line of resistance is approximately 
the true one which lies nearest to the axis of the arch-ring as 
determined by the method of least squares." 

In this discussion the following definitions are adopted : — 

i°. The axis of the arch is a plane curved line passing 
through the centres of gravity of all its normal sections. 

2°. The plane of this axis is called the plane of the arch. 

3 . The axial layer of the arch is a cylindrical surface per- 
pendicular to the plane of the arch, and containing its axis. 

4 . A section normal to the axis is called a cross-section. 

5°. The length of the axis between two sections is called 
the length of arch between the sections. 

The loads may be single isolated loads, or they may be 
■distributed loads. 

We shall, in this discussion, assume in the plane of the arch 
a pair of rectangular axes, OX and O Y, positive to the right and 
upwards respectively. 

We will, then, assuming any point on the axis of the arch 
before the loads are applied, call x, y, the co-ordinates of that 
point, s the length of axis from some arbitrary fixed point, <£ the 
angle made by the tangent line at that point with OX, r the 
radius of curvature of the axis at that point, x -\- dx, y -f- dy, 
s + ds> and <j> -f- d§, the corresponding quantities for a point 



828 



APPLIED MECHANICS. 




yery near the first before the load is applied ; also we will denote 

by 7) the perpendicular distance of 
any fibre from the axial layer, by s+ 
the length of arc measured to that 
point where this fibre cuts the cross- 
section through (x, y) } and s v + 
dsr, the length of arc measured on 

this fibre to the next cross-section r 

FlG 2 9 x - so that ds will be the distance apart 

of the cross-sections measured on the axis, and ds n on the other 
fibre. All this is done before the 
load is applied, and is shown in 
Fig. 291 ; while the changes brought 
about by the application of the 
loads combined with change of tem- 
perature are denoted by A's, and 
shown in Fig. 292. Thus, x, y, s, 
and (p become respectively x + 
Ax, y -f- dy, s -f- As, and -(- A(f>, 
Now the course we are to fol- 
low in the discussion is, to imagine 
a cross-section dividing the arch into two parts, and to impose 
the conditions of equilibrium between the external forces acting 
on the part to one side of the section, and the forces exerted 
by the other part upon this part at the section. These latter 
forces may be reduced to the three following : — 

1°. A normal thrust T x uniformly distributed over the sec- 
tion, the resultant acting at the centre of gravity of the section. 
2°. A shearing-force S x at the section. 

3 . A bending-couple at the section ; this comprising a 
stress varying uniformly from the axial layer, and amounting 
to a statical couple, tension below, and compressions above, the 
axial layer. 

Moreover, (1) and (3) combined amount to a uniformly vary. 




GENERAL THEORY OF THE ELASTIC ARCH. 829 

ing stress, the magnitude of whose resultant is T xy its point of 
application not being at the centre of gravity of the section ; 
this sort of composition having been already exhibited in the 
case of the short strut (§ 207). 

Now, let r be the radius of curvature of the axial layer at 
the section ; and we have, from Fig. 291, by similar sectors, 

dsy = ds + r)( — d<f>) = ds — r)d<f>. (1) 

But 

r(-dcf>) = ds /. ^=-1 

ds r 



^ = 4+2) = ^l±?). (,) 



Now, if the loads are applied, and the changes take place 
that are indicated in Fig. 292, we shall have, by suitable sub- 
stitutions in (1), 

d(s v + Aj„) = d(s + As) - v d(<j> -f A0) ; (3) 

and, combining this with (1) and (2), we obtain 

dAs v _ fdAs _ dA<f> \ r , v 

dsy, \ ds ds jr -f- rj 

Now, the change of length of fibre from ds v to d(s v + As v ) is 
due to two causes : (1) the change of temperature, (2) the stress 
acting on the fibre normal to the section. 

Let e = co-efficient of expansion per degree temperature, 
r = difference of temperature, in degrees. 
Pr, = intensity of stress along the fibre at section. 
E = modulus of elasticity of the material. 
Then 

€T -& = — ^ = (— - /^\ r (xC\ 

E ds^ \ds ' ds )r + rf y:} ' 

Hence, solving for p^ we have 



>-^-f)^ + ^> <« 



83O APPLIED MECHANICS. 

— — -I... - ^ 

this being the expression for the stress per square inch on the 
fibre whose distance is yj from the axial layer. 

Hence we shall have, by summation, if elementary area = 
dA, 

[VA<£ rqdA d&s^ rdA ~| , x 

and for the moment M x we have, by taking moments about the 
neutral axis of the section (i.e., horizontal line through its cen- 
tre of gravity), 

rd&dt^rrfdA d&s _ rndA 1 

m x = % p ^a = ^L^ 2 r+", - nr*F+- v + *H- (8) 

Let IdA = A, r% ^-^- = fi, and observe that SqdA — o, 

since the axis passes through the centre of gravity of the sec- 
tion, and we have 

r -f- rj r -{- r) r 

rf ™k4 1 1 _ w 2 ^ O 

2^4 - -2^ 4- -S-^t— = -4 + -< 



r -\- rj ■ r r r -f- rj r * 

Making these substitutions, we have 

T x /dk<f> 1 dks\Cl (dks \ 

E = ~\W + ~r ~d7jr ~ \di ~ €T ) A > 

_ / d&<j> dAs\Q 

= \ r ~dT + ~di]7 



E 



Hence, solving for — i and — ?, we have 
ds ds 

, --^+^U ! + — K, ( 9 ) 









GENERAL THEORY OF THE ELASTIC ARCH. 83I 

Now, from Fig. 292, we have 

d(x + Ax) = d(s + As) cos (<f> -f A<p), 
d(y + Ay) = d{s + Ax) sin (<j> + A0); 

but, if we write cos A<£ = 1, and sin A<£ = A0, 

*£c dy 

cos(<f> + A</>) = cos<£ — A<£sin<£ = -=- — A<f>-~t 



Hence 



dy dx 

sin(<£ + A<£) = sin<£ -f A<fiCos<j> = -y + A<£-r-« 

</A* = -A<^ + — dx - (^dyA<f>\ 
ds \ ds ) 

dAy = +A<}>dx + —dy + (^dxAcf>\. 
ds \ ds / 

or, omitting the last terms, and integrating, 

A* = -fA<j>dy + fYdx, (n) 

Ay = fA<f>dx + JYdy; (12) 

and, integrating (9) and (10), 

A* = /Yds, (13) 

A<f> = fXds. (14) 

In these four equations we have 

Ax = horizontal deflection due to the loads, 
Ay = vertical deflection due to the loads, 
As == change of length of arc due to the loads, 
A<f> = change of slope due to the loads. 

The three equations which we shall have occasion to use are 
(11), (12), and (14), and if we make the integrations between 
the limits x and o, they become, by changing their order, 



A<t> = A<t> + fxds, (15) 



832 APPLIED MECHANICS. 

Ax=— f J<pdy + f*Ydx, (16) 

4y= f*~J<f>dx+ f*Ydy, (17) 

e/*=o t/x=o 

where d(f) is the change of slope for x = o. 
If, now, we write 

_ . J/* M x T x . 

(19) 







we shall have 






r 



(20) 

K=-^ + er; (21) 

or if we neglect the effect of temperature, we may write 

X = M It (22) 

Y=-P l . (23) 

Moreover, we may with very little error substitute the 
moment of inertia / for £1 in the value of M lt i.e., writing this 

_. M x M x T x 

§ 272. Manner of using the Fundamental Equations to 
Determine the Stresses in an Iron Arch. — In order to be 
able to determine the stresses in all the members of an iron 
arch with any given loading, we need to determine the three 
quantities T x , S Xf and M x for each section. 

Now, if we let R x represent the thrust at the section, we 
shall have 



DETERMINA TION OF STRESSES IN AN IRON ARCH. 833 



R x =\lT x * + S x *; (1) 

and, if we let H x and V x represent the horizontal and vertical 
components of R x respectively, we have that we need to deter- 
mine the three quantities H x , V x , and M x for each section. 

Let us suppose the arch to be subjected to vertical loads 
only, and let 

H = horizontal component of thrust at all points, 
V = vertical component of left-hand supporting force, 
V x = vertical component of right-hand supporting force, 
M = bending-moment at left-hand support, 
M' = bending-moment at right-hand support. 
Assume origin of co-ordinates at left-hand support, and 
x -f- to the right, and y -f- upwards, and impose the conditions 
of equilibrium upon the forces acting on the part of the arch 
between the section and the left-hand support ; then we have, 
if W\s any one load, and a the x of its point of application, 

H x = H, (2) 

V x = V- X> X W 9 (3) 

M x = M + Vx - By - % Q x W(x - a). (4) 

Hence it is plain that the three quantities which we need 
to determine are H, V, and M. 

Now these are also the three unknown quantities which will, 
by suitable reductions, become the three unknown constant 
quantities in equations (11) to (14). The determination of these 
three quantities requires three conditions ; what these condi- 
tions are depends upon the manner of building the arch, as will 
be seen from the following three special cases : — 

Case I. — Let the arch be jointed at three points, viz., the 
two supports, and one other point whose co-ordinates are x = 
x x and y = y x . Then we know, that, for all points where 
there is a hinge, there can be no bending-moment. Hence 



834 APPLIED MECHANICS. 

M = o, M' = o, and M Xi = o, 

which are the three required conditions ; and, if these be im- 
posed, it is easy to obtain H xi V xy and M xt for every section. 

CASE II. — Let the arch be jointed only at the ends. Then 
M = M' = o gives us two conditions : and for the third we 
have Al = o ; i.e., if we put / for x in equation (16), § 271, 
after having made the integrations, we have the third equa- 
tion, as this expresses simply the condition that the sup- 
ports remain at the same horizontal distance apart after the 
load is put on as before. With these three conditions we can 
determine H x1 V x1 and M x for all sections. 

Case III. — Let the arch be fixed in direction at the ends. 
We must now have three conditions. These will be as follows: — 

1°. Al = o ; i.e., the supports remain at the same horizontal 
distance apart after the load is applied as before. 

2°. Ah = o (h being the difference of level of the sup- 
ports) ; i.e., the supports remain at the same vertical distance 
apart after as before the load is applied. 

3 . A0 l = o; i.e., the tangents at the ends make the same 
angle with each other after as before the load is applied. 

The value of A<p x is obtained by integrating (15), § 271, and 
then substituting / for x, or h for y y observing that A<p = o. 

The value of Al is obtained by integrating (16), § 271, and 
then substituting / for x. 

The value of Ah is obtained by integrating (17), § 271, and 
then substituting / for x, or h for y. 

In this case, if we neglect the effect of temperature, write 

/ for n, omit all terms containing -, and also neglect T x in 

(15), (16), and (17) of § 271, we shall obtain by making one 
integration, 



DETERMINATION OF STRESSES IN AN IRON ARCH. 835 



AX = SeT**' (5) 

A< f= /§<*• (7) 



While it has often been proposed to use these as approxi- 
mately true, nevertheless the degree of approximation is too 
coarse to render them suitable to use in practice. 

CIRCULAR ARCH, UNIFORM SECTION, AND VERTICAL LOADS. 

We will next deduce expressions for A<f>> Ax, and Ay, for a 
circular arch of constant cross-section and loaded vertically, 
and thence deduce the equations from which to determine the 
three quantities M, M' , and H in any such case, and also the 
expression for the horizontal thrust in an arch hinged at the 
two springing-points, and symmetrical in form. We will write 
in place of £1 the moment of inertia /, and will neglect terms 

containing -, in equations (15), (16), and (17), but will not 

T 

neglect T x . 

Take the origin at the left-hand springing-point, and the 
axis of x horizontal. , 

Observe that if represent the angle the tangent line to 
the arch at the point (x, y) makes with the axis of x, it also 
represents the angle subtended by the radius drawn through 
the point (x, y) with the vertical radius, i.e., that through the 
crown. 

Let O be the value of at the origin, and let a be the 
value of at the point of application of any concentrated load 
IV, the co-ordinates of this point being (a, b). 



&$6 APPLIED MECHANICS. 

Let the co-ordinates of the centre of the circle be 

g = r sin O , — k = — r cos O ; 

of the crown be g = r sin O , / = r — r cos O ; 

of the point of ap- 
plication of IV be a = r(sin O — sin a), b — r(cos tf — cos O ); 

and of any point 

on arch, x = r(sin O — sin 0), jy = r(cos — cos O ). 

The following is a list of relations which can be easily 
proved, and which are needed for use in the work that follows 
them. 

g — x = r sin 0; 7^ = V x sin + H cos 0; 

*+J = '«■*; . *-„ = *■+ Fi-iZy-i^*- a); 

jc — # = r(sin <z — sin 0) ; 

^-^ = rsin«; ^ = F _i^ ; 

y — d = r(cos — cos «) ; ° 

* = - "*0J it/* = M + VI- He- 2W(l- a); 

dx == — r cos 0^0; ° 

^V = — r sin 0^/0; 
where J/' = bending-moment at right-hand springing-point. 

Also 

M T 

By making the substitutions indicated, and also the inte- 
grations, we obtain from (15), (16), and (17) the following: — 

A<t> = Acf> o + -^ { (0 O -<fi)(M+ Vr sin Q + Hr cos O ) 

X 

— Vr (cos — cos O ) — «£Tr (sin O — sin 0) -f- 2 Wr (cos 

o 

X 

— cos ar) — 2JVr (a — 0) sin <*}, (8) 

o 
f 

Ax — etx — ^-^ I r(sin a O — sin 2 0) + Zf [sin O cos O 

— sin cos + (0 O — 0)] — ^ ^(sin 2 a — sin 2 0) ( —yA<p Q 

- -^1^+ ^ sin O + -ffr cos O ) [(0 O - 0) cos 



DETERMINATION OF STRESSES IN AN IRON ARCH. 837 

Vr Hr 
— (sin O — sin 0)] (cos — cos0 o )' [ — (0 O — 0) 

-f- cos (sin O — sin 0) + sin O (cos — cos O )] 

x x 

-\- ir2 W (cos — cos a)* — r cos <p2W (a — 0) sin a 

o o 

-{- r^ W sin a (sin a — sin 0) } ; (9) 

Ay = ety - -^ {ZT (sin 8 O - sin 3 0) - V [sin O cos O 

X 

— sin cos — (0 O — 0)] — 2 W [sin cos — sin a cos « 

o 

+ (a -</>)]] + x4<P +£j{(M+ Vrsm<f> 

+ Hr cos O ) [(cos — cos O ) — (0 O — 0) sin 0] 

Vr 

— fffjr (sin O — sin 0) 8 [— cos O (sin O — sin 0) 

— sin (cos — cos O ) + (0 O — 0)] + sin 0^ W>V (a 

o 

X 

— 0) sin a — 2 Wr sin a (cos — cos a) 

o 

— ^2 Wr [cos « (sin a — sin 0) + sin (cos — cos a) 

-(«-*)]}■ (10) 

We will next write out the same values as applied to the 
right-hand springing-joint of a symmetrical arch. 

In this case we have the value of for the right-hand end, 
or r equal to — O ; and if we make this substitution, observing 
that x becomes / and y becomes zero, and if we substitute for 
Fthe value 

# - M , 2Wr (sin O + sin a) 
r sin O r sin O 

then we obtain the following : — 

A<t>, ■= J0 O + -^ { (M' + M) O - 2^ (sin O - O cos O ) 

— %2Wr\_2<x sin ar — 20 o sin O + 2 (cos a — cos O )}; (11) 



838 APPLIED MECHANICS. 

4/= erl- ^{^(20 o +2sin0 o cos0 o )+i^(sin> o -sin'a)f 

+ --gj I {M* + M) (2 sin O - 2 o cos O ) - Z^( 4 Q cos* O 

— 6 sin O cos O + 20 o ) -\- 2 Wr [2a cos O sin « 

o 

+ 2 cos O cos a — 20 o sin O cos O + sin 3 O — sin 9 a 

— 2cos a o ](; (12) 

Ac = lA<f> + ^j\{M' + M) ( 2 o sin O ) - *Hr (sin 3 O 

— O sin O cos O ) — (M - M') (cos O - — -^- 

— \^,Wr [20 o (—-7 2 sin 9 O J -f 2a (2 sin O sin a — 1) 

— 4 sin O cos O + 2 sin a cos O — 2 sin <z cos a 
+4 cos « sin O ] f - -^ { (JT - JO (cos * - ^j 

/ r 20 o . "1 ) 

-\-\'2Wr\ - — °— sin a — 2a — 2 sin a (cos « — cos O ) >■ . (13) 

SPECIAL CASES OF SYMMETRICAL ARCHES. 

i°. Three -hinged Arch. — In this case we do not need these 
equations to find the horizontal thrust : the proper ones can 
be used subsequently if we wish the deflections or slopes. 

2°. Arch hinged at the two springing-points. — In this case 
M = M' == o ; and by making these substitutions in (12) and 
solving for H, we obtain 

2W[2a cos O sin a-\-2 cos O cos a—2<f> Q sin O cos O 

+ sin" O — sin 3 a — 2 cos 3 O ] 

/ 2FI 

--^2W(sin> O - sin 3 «)--=■ (J/ - er/) 

jr= ^ _ r - .(14) 

40 o cos 3 o — 6 sin0 o cos0 o +20 o +— 2 (20 o +2 sin0 o cos0 o ) 



DETERMINATION OF STRESSES IN AN IRON ARCH. 839 

» — — 

This formula gives the thrust when the value of Al is known, 
i.e., the amount of relative yielding of the supporting points. 

If the abutments do not yield at all, then Al = o, and that 
term should be omitted from the numerator ; so, also, if we 
neglect a consideration of the temperature, then erl vanishes 
in addition. 

Formula (14) gives the thrust for a set of concentrated loads, 
each equal to IV. 

For a distributed load, we should substitute for W t wdx y 
and integrate between the proper limits, w being the intensity 
of the load per unit of horizontal length, and being constant or 
variable according to the distribution of the load. 

The formula for the thrust in the case where the load is 
uniformly distributed horizontally (i.e., when w is a constant) 
and when it covers the entire arch will now be given, but will 
not be worked out here, as it is easily obtained from (14). 

In this formula the letters have the same meanings as 

heretofore, and we use also A x = I ydx = area of segment of 

arch; and let m = -— t . 
Ar 

The formula is as follows : — 

(1 - — j) - + k\ 2A X +- sin 20 o — O (Z'+r 2 cos 20 o ) \ 

(^ + 2r')0 o - 3 £/+ m (2Scj> +kl) ' K 5) 

In a similar way formulae are easily obtained for the thrust 
when half or a quarter, or some other portion of the arch, is 
loaded. 

3 . Arch with no hinges. — In this case, if we know A<p, A I, 
and Ac, or if these are zero, we can obtain from (11), (12), and 
(13) the three quantities M, M' y and H, and then the solution of 
the arch follows. 

This will not be done here, however, as the arches usually 
built are of the other kinds. 



84O APPLIED MECHANICS. 

" • - ■ t. 

EXAMPLES. 

i. Given a semicircular arch jointed at each springing-joint and at 
the crown, radius r. Trace out the effect of a single load W acting 
upon it at the extremity of a radius making 45 ° with the horizontal. 

Solution. 

The presence of three joints gives us the bending-moments at each 
of these joints equal to zero, the co-ordinates of these joints being 
respectively (o, o), (r, r), and (2?-, o). 

Hence, using equation (4), we obtain 

i°. M = o, 

2 . Vr — Hr — ^(0.7071 1 r) = o, 

3 . V(*r) — ^(1.70711/-) = o. 

Solving, we have, therefore, 

V = 0.85355 W = left-hand supporting-force, 
and 

H = 0.14645 W — horizontal component of thrust 

Hence V x — 0.14645^= right-hand supporting-force. 
Hence, for a section whose co-ordinates are (x,y), 

x < 0.29289*-, V x = 0.85355 W; 

x > 0.29289/-, V x = — 0.14645 W. 

Hence equation (1) gives, for 



x < 0.29289?-, R x = ^(0.85355)* + (0.14645) 2 

= 0.86602 Wy 

x > 0.29289;-, R x = JfV(o.i4645) 2 + (0.14645) 2 

= 0.2071 1 IV. 

Now, the angle made by R x with the horizontal is, for 

x < 0.29289,-, a, = tan-*(^g|) = 8o° 15' 51", 

* > 0.29289,-, a, = tan-gggj) = 45°. 



TRUE LINE OF RESISTANCE IN A STONE ARCH. 841 

Knowing, now, the angle made by R x with the horizontal, we can 
find, for the point (x, y) , the angle made by a tangent to the circle with 

the horizon, or a, = tan~ x f J. Then resolve R x into two com- 
ponents, respectively tangent to the arch and normal to it at the point 
x, y, and the tangential component is the direct thrust T X} while the 
normal is the shearing-force S x . 

Then, for the bending moment, we have, from (4), 

x < 0.29289?-, M x — 0.85355 Wx — 0.14645 Wy ; 

x > 0.29289?-, M x = 0.85355 Wx — 0.14645 Wy — W{x — 0.29289^ 

Hence we determine the direct thrust, the shearing-force, and the 
bending-moment at any section, and can hence obtain the stresses at all 
points. 

2. Given the same arch with a load W distributed uniformly over 
the circular arc, find stresses at all points. 

3. Given the same arch jointed only at the two springing-points, 
find stresses at all points. 

§ 273. Position of True Line of Resistance in a Stone 
Arch. — The proof will now be given of the proposition already 
referred to in regard to the position of the true line of resist- 
ance ; viz., — 

" For an arch of constant section, that line of resistance is 
approximately the true one which lies nearest to the axis of the 
arch-ring, as determined by the method of least squares." 

Proof. — If we denote by y the ordinate of the axis of the 
arch for an abscissa x> and by > that of the line of resistance 
for the same abscissa, then i*. — y is the vertical distance be- 
tween the two curves for abscissa x. Now, the condition that 
the line of resistance should be as near the arch-ring as possi- 
ble, is, that the sum of the (jjl —y) 2 shall be a minimum, or 

/(/a — y) 2 ds — minimum. (1) 

But (x, fi) are the co-ordinates of the point of application of the 



842 APPLIED MECHANICS. 

actual thrust, and hence (fx — y) is the distance of the point at 
which the resultant thrust acts from the centre of gravity of 
the section. Hence we have 

Hence (1) becomes 



/(f)' 



ds = minimum. (2) 



But H is constant for the same line of resistance, though it 
varies for different lines : hence we can place H outside of the 
integral sign. Hence we may write 

M x 2 ds = minimum. (3) 



u == — \ 1 



Now, from (4), § 272, we have 

M x = M + Vx - Hy - 2 x W(x - a) = <f>(M, V, ff); 

My V, and H being constants for the same line of resistance, 
but varying for different lines. Hence, by differentiating (3), 
we have 
du du dM x 2 Cr 7 C^ , , c' 

du du dM, 



dH dM x dH 



= -2i7-3 \M x 2 ds - 2H- 2 \M x yds 

= - B ~ l \jj 2 J M * ds + 2 JiS MxydS \ = °* 

But the first term must be very small : hence we may write ap- 
proximately, 

fM x yds = o. (6) 

Now, the three expressions (4), (5), and (6) are identical with 
(5), (6), and (7) of § 272 ; and the conditions that these shall 
be zero are, with the degree of approximation there stated, the 



DOMES. 



343 



conditions that hold in the case of an arch fixed in direction at 
the ends. Hence it follows that the condition that the line of 
resistance shall fall as near the centre of the arch as possible is 
the condition which, in an elastic arch fixed in direction at the 
ends, gives us its true position. Hence it would seem that 
the most probable position for the true line of resistance is the 
nearest possible to the axis of the arch. 

This is the conclusion reached by Winkler ; and a more 
detailed discussion of the matter is to be found in an article 
by Professor Swain in "Van Nostrand's " for October, 1880. 

§ 274. Domes. — The method to be used for determining 
the stability of a dome differs essentially from that used in the 
case of an arch, for there is no thrust at the crown in a dome. 
Indeed, the most general case is that of the dome open at the 
top . we will, therefore, consider this case first in studying the 
action of the forces required to preserve equilibrium. 

Fig. 293 shows a meridional section of an open dome, 
pose that this dome had been 



Sup- 



rP 



entirely built, except the up- 
per ring-course of stones, rep- 
resented by LKGH. Then, 
suppose that one of the stones 
only of this course were placed 
in position without any auxil- 
iary support, its own weight 
would evidently overturn it, 
since the line ab, along which 
the weight acts, does not cut 
the joint ; but, if the whole 
ring-course is put in place, the 
stones 'keep each other in po- 
sition. The way in which this 
is accomplished is as follows : 
they press laterally against each other; and the resultant of the 




^ 



Fig. 293. 



844 APPLIED MECHANICS. 

pressures exerted upon the two lateral faces of any one stone 
by the other stones of the course is a horizontal radial force, 
which, combined with the weight of the stone, gives, as the 
resultant of the two, a force which cuts the joint between G 
and H. Moreover, sufficient pressure will be developed to 
accomplish this result, as a failure to reach the result will only 
increase the pressure upon the lateral faces. 

Moreover, if, when sufficient pressure has been developed 
to bring the resultant of the weight of the stone and the above- 
described horizontal radial force within the joint, it should 
make an angle with the normal to the joint greater than the 
angle of friction, the tendency of the stone to slide will increase 
the lateral pressure, and this in turn will increase the outward 
horizontal force till the angle made by the resultant with the 
normal to the joint is no greater than the angle of friction of 
the material of the voussoirs. 

This will be made plain by reference to the figure (Fig. 
293), where ab represents the weight of the stone HLKG, and 
where 0/3 is perpendicular to HG and Oy is drawn so that yG/3 
= <f>, the angle of friction. Now, since ab produced passes out- 
side of HG, horizontal thrust must be developed. And, more- 
over, were only sufficient horizontal thrust furnished to make 
the resultant cut HG at G, the angle between this resultant 
and the normal to the joint would be greater than <f> ; there- 
fore we proceed as follows : assuming the horizontal thrust to 
act through L, the upper edge of the stone, we lay off from b, 
the intersection of the horizontal through L with a vertical line 
drawn through the centre of gravity of the stone, the weight ab 
to scale, then from b draw be parallel to Oy, and draw through 
a a horizontal line to meet be. Then will qe be the horizontal 
force that will be furnished by the other stones of the course 
to keep this stone in place ; and the pressure upon joint HG 
is be, and acts at the intersection of be and HG. 

Now prolong be to meet the vertical drawn through the 



DOMES. 845 



centre of gravity of the next stone, HGFE, at d. Combine it 
with the weight of this stone ; this is done by laying off de = be, 
and from e drawing ef vertical, and equal to the weight of 
FHGE. The resultant fd makes an angle with the normal to 
FE greater than <j> : hence draw 0& perpendicular to FE and 
Oe, so that c(98 = <J> ; then f rom g, the intersection of <^"with a 
horizontal line through H, the top of FHGE, lay off gs = df y 
through g draw gh parallel to Oe, and through s draw sh hori- 
zontal. Then is sh the horizontal thrust that will be furnished 
at H to keep the stone HGEF in place ; and kg is the pres- 
sure upon joint FE, and acts at the intersection of FE with kg. 

Next, prolong kg to meet the vertical through the centre of 
gravity of stone FEDC at k ; lay off kl — gh, and from / lay 
off Im = weight of stone FEDC ; draw km, which cuts the 
joint within the joint itself, and needs no horizontal thrust to 
bring it inside ; hence mk is the pressure on joint DC. 

Then draw mk to meet the vertical through the centre of 
gravity of ABCD at n, and lay off no = km ; draw op = weight 
of ABCD, and draw pn, which will be the pressure on the joint 
BA. 

It is necessary, for stability, that all these forces should cut 
the joint inside of the joint if the stones are reckoned incom- 
pressible ; or we may adopt the middle third, or other limits, as 
our criterion of stability. 

As long as it is outward thrust that is required to produce 
stability, it is possible to furnish it ; but, if we should reach a 
joint where inward thrust would be required, this could not be 
furnished, and the dome would be unstable. Moreover, the 
resultant pressure on the springing gives us the pressure ex- 
erted upon the support of the dome ; and it must not cut any 
joint of the support outside of that joint, as otherwise the sup- 
port would not stand. 

In determining the numerical value and direction of this 
pressure on the support, we may either construct it graphically, 



846 APPLIED MECHANICS. 

or we may compute it as follows : (i°) Compound all the ver- 
tical forces, i.e., the weights, and find the magnitude and line 
of action of the resultant of these. (2 ) Compound all the 
horizontal forces, and find the magnitude and line of action of 
their resultant (in this case the horizontal forces are two ; viz., 
ac applied at L, and sh applied at H) ; then compound these two 
resultants. The graphical and analytical method should check 
if no mistake has been made in the work. 

In the above calculation, it has been assumed that the figure 
represents the portion of a dome included between two merid- 
ional planes. 

If we desire to ascertain the pressure exerted upon the 
lateral face of the stone by its neighbors in the same ring- 
course, we only need to know the angle made by the two 
meridional planes containing the lateral faces of the stone in 
question, then resolve the horizontal thrust upon that stone 
into two equal components, which make with each other an 
angle equal to the supplement of the angle of the planes ; i.e., 
resolve the outward horizontal thrust into two components 
normal to the lateral faces. 

In regard to the assumption that the outward thrust acts at 
the top of the stone, it should be said that this is Schefrler's 
custom, his reason being that less thrust will be required if he 
assumes it at the top than if he assumes it nearer the middle. 
The true position of this thrust is probably much nearer the 
middle of the stone. 

An example will next be solved, giving Schefrler's method 
of working. 

Example. — Given the dome shown in the figure, sur- 
mounted by a lantern at the top ; determine whether it is 
stable, and what should be the thickness of the support in order 
that the resultant pressure may not pass outside any joint of 
the pier. 

The dimensions are as follows : — 



DOMES. 



847 



Diameter of outer vertical circle = 20 feet. 

Diameter of inner vertical circle =18 feet. 

Angle made by springing- radius with vertical = 75 ° = 
angle A OB. 

The inner edge of the upper 
voussoir subtends 18 on the 
lower circle ; the width of the 
load of the lantern is 0.6 ; 
the voussoirs below that, each 
subtend 18 . 

Assume 36 stones in a hori- 
zontal course. The width of the 
lowest will, then, be 1.5 1 ; the 
width of the others are deter- 
mined from their lever arms. 

Given height of pier = 8 
feet. 

Height of the centre of the 
sphere above base of pier = 8' 
— 10 sin 15 = 5.41'. 

The figure may be taken to 
represent the portion of the 
dome included between two ver- 
tical planes passing through the 
axis of the dome : hence it 
shows one vertical series of 
stones. 

We first construct a table 
giving the weights of the differ- 
ent voussoirs with any superin- 
cumbent load, their centres of gravity, and the moments of 
their weights about an axis passing through 0, and perpen- 
dicular to the central plane of the portion shown ; and we 
so choose our unit of weight that the volumes of the 




Fig. 294. 



848 



APPLIED MECHANICS. 



voussoirs shall represent their weights, 
as follows : — 



The work is arranged 



Elementary Forces. 


Horizontal Forces. 


(1) 


(2) 


(3) 


(4) 


(5) 


(6) 


(7) 


(8) 


(9) 


I 

2 
3 

4 


Area of 
Lateral Face. 


Thick- 
ness. 


Product. 


Lever 
Arms. 


Moment. 


Hori- 
zontal 
Forces. 


Lever 
Arms. 


Moment. 


O.6X6.680 
2.985 
2.985 
2.985 


0-53 
O.74 
I.23 

I-5 1 


2.124 
2.209 
3.672 

4-5°7 


3-°7 
4-75 
7-05 
8.68 


6.521 
IO.493 
25.888 
39.121 


I.74 
I.26 
I.32 


9.60 

9-33 

7.78 


16.IO4 
II.756 
IO.273 


- 


- 




12.512 


- 


82.023 


4.32 


- 


38.730 



Column (1) contains the numbers of the voussoirs, counting 
from the top. 

. Column (2) contains the areas of the lateral faces of the 
stones shown in the figure. For the three lower stones, the 
area of a ring subtending 18 at the centre, and of the dimen- 
sions given, is calculated. For the first, the height is 6.6S and 
the width 0.6. 

Column (3) contains the thicknesses of the voussoirs ; i.e., 
the length of arc between their two lateral faces measured on a 
horizontal circle through the centre of gravity of the voussoir, 
which is here taken at the middle point of the arc subtended 
by this voussoir on its middle vertical circle, i.e., one which 
has a radius 9.5 feet. 

Hence, the thickness of the lower stone being 1.5 1 feet, that 
of the others will be 



(I,5,) SI = °- 53, 



(i - 5i) s3 - °- 74 ' 



7-°5 



^m = "* 



DOMES. 849 



Column (4) gives the weights of the voussoirs and their 
loads : it is obtained by multiplying together the numbers in 
columns (2) and (3). 

Column (5} gives the distances of the centres of gravity of 
the different voussoirs from the axis of the dome : it may be 
determined graphically or by calculation. 

Column (6) gives the moments of the weights about a hori- 
zontal axis through O perpendicular to the central plane of this 
series of voussoirs. The graphical Construction for determin- 
ing the horizontal thrusts required is next made, and the results 
are recorded in column (7). It will be seen that no thrust is 
required on voussoir No. 4. 

Column (8) contains the lever arms of these forces about 
the same axis. 

Column (9) contains their moments about the same axis. 

The construction thus far has shown no case where horizon- 
tal tension instead of horizontal thrust is required to cause the 
thrust on any joint to pass within the joint : hence thus far the 
dome is stable ; and the question comes next as to what should 
be the width of the pier in order that the line of resistance, if 
continued down, may remain within it. 

For this purpose we proceed as follows : — 

Let t = thickness required. 

Let breadth be equal to that of the lowest voussoir. 

Height = 8 feet. 

Take moments about the outer edge of the base of the pier. 

We shall then have, — 

i°. Moment of vertical load on dome, and of weight of dome 
sector about inner edge of springing, = 

(12.512) (8.69 - 6.56) = 26.52. 

2°. Moment of same about outer edge of springing of pier = 
26.52 + (12.512)/. 



850 APPLIED MECHANICS. 

3 . Moment of horizontal forces about the same axis = 

38.730 + (4-32) (5-4i) = 62.101. 

4 . Moment of weight of pier about outer edge = 

{ 8(1.5 i)/|J/« 6.04/2. 
Hence we have 

6.04/ 2 -f 12.51/ -f- 26.52 = 62.101 

.\ t 2 + 2.07/ = 5.89 .*. / = 1.60 feet. 

This is the thickness required in order that the line ot 
resistance may remain within the lower joint. 

If, on the other hand, while pursuing the same method with 
the dome itself, we require that the line of resistance shall 
remain within the middle third of the pier, we take moments 
about a point in the springing of the pier at a distance § / from 
its inner edge, we should then have 

\P +1(2.07)/= 5.89 

,\ t 2 + 2.07/ = 8.84 .-. / = 2.10 feet. 

On the other hand, we could proceed in a similar way to 
the above, if we desired to keep the line of resistance in the 
dome within the middle third, by merely assuming the horizon- 
tal thrusts to act at two-thirds the thickness of a joint from the 
lower edge, and using a point two-thirds the thickness from 
the top, instead of the lower edge, as the lower limiting-point 
for the pressure to pass through. 

This will not be done here, however. 

Example. — As an example, St. Peter's dome will be given, 
with the dimensions as given by Scheffler reduced to English 
measures. The dome consists in its upper part, as will be 
evident from the figure, of two domes ; the lantern resting on 



DOMES. 



8 5 I 



the two is assumed to have one-third of its weight resting on 
the upper, and two-thirds on the 
lower dome. 

Diameter of dome = diameter 
at the base = 144 feet. 

Up to a point 28.48 feet above 
the point £7 it is formed of a single 
dome 11.84 f eet thick. In its 
upper part, on the other hand, 
it is composed of two domes 
whose normal distance apart is 
5.15 feet; the exterior having a 
thickness of 2.56 feet, and the 
inner of 4.13 feet at the top and 
5.15 feet at the springing. At 
the top of these two domes is an 
opening 12.24 feet radius, sur- 
mounted by a cylindrical lantern. 
The magnitude of the load of the 
lantern on the dome is repre- 
sented on the figure by 1.82 feet 
width and 56.66 feet height. 

Height of the entablature 
ABCD — 23.69. 

Width of ABCD normal to 
plane of paper = 1.02 feet. 

Thickness of ABCD = 10.30 
feet. 

Divide the exterior dome into 
nine parts, the interior into eight 
of a uniform circumferential width 




Fig. 295. 



of 10.08 feet, except the 
first, which has a width of only 1.82 feet. 

Determine whether this thickness of ABCD is sufficient to 
keep the line of resistance within joint AB. 



852 APPLIED MECHANICS. 



CHAPTER X. 
THEORY OF ELASTICITY, AND APPLICATIONS. 

§ 275. Strains. — When a body is subjected to the action 
of external forces, and in consequence of this undergoes a 
change of form, it will be found that lines drawn within the 
body are changed, by the action of these external forces, in 
length, in direction, or in both ; and the entire change of form 
of the body may be correctly described by describing a suffi- 
cient number of these changes. 

If we join two points, A and B, of a body before the exter- 
nal forces are applied, and find, that, after the application of 
the external forces, the line joining the same two points of the 
body has undergone a change of length A(AB), then is the limit 

of the ratio , as AB approaches zero, called the strain of 

the body at the point A in the direction AB. 

If AB -\- \{AB) > AB, the strain is one of tension ; whereas, 
if AB -(- &(AB) < AB, the strain is one of compression. 

In order to study the changes of form of the body, let us 
assume a point within the body when there are no external 
forces acting, and let us draw through this point three rectangu- 
lar axes, OX, OY, and OZ, and assume a small rectangular 
parallelopipedical particle whose three edges are OA, OB, and 



STRAINS. 



S53 



OC, and let us examine the form of this particle after the 

loads are applied ; it will be 

found that the edges OA, OB, 

and OC will be of different 

lengths from what they were 

before, and that the angles 

AOB, AOC, and BO (Twill no 

longer be right angles, but 

will differ slightly from 90 . 

Let the parallelopiped oabc— 

g def represent the form and 

dimensions of the particle 

after the external forces are 

applied. Then we shall have, 

if a x , ey, and c 2 represent the strains in the directions OX, OY t 

and OZ respectively, that 

v •*. r P ro J- °g on OX — OA ~ . , 

e x = limit of - — - — - — 7V - A as OA approaches zero, 




Fig. 296. 



limit of 



OA 

proj. og on OY — OB 
OB 



as OB approaches zero, 



1- •*. rP ro J- °g on OZ — OC ~~ , 

:, = limit of K. — t — * — — — as OC approach 



es zero. 



In the figure, * x and c 2 are tensile strains, and e y is a com« 

pressive strain. 

But these strains do not represent completely the distortion 
of the particle ; for the plane CEGD has slid by the plane 
OABF through the distance oc If the distance apart of these 
planes being OC, and the plane halfway between the two has slid 
just half as far, so that the amount of shearing, or the shearing- 
strain of planes parallel to XOY in the .direction OX, may be 

represented by — -^ = ~ nearly, or the distortion divided by the 
OC cc t 



854 APPLIED MECHANICS. 

distance apart of these planes. This, moreover, is the tangent 
of the angle occ x , or the tangent of the angle by which aoc 
differs from a right angle. 
If, now, we let 
y zx = shearing-strain in a plane perpendicular to OZ in the 

direction OX, 
y zy = shearing-strain in a plane perpendicular to OZ in the 

direction OY, 
y yx = shearing-strain in a plane perpendicular to O Y in the 

direction OX, 
y yz = shearing-strain in a plane perpendicular to O Y in the 

direction OZ, 
y xz = shearing-strain in a plane perpendicular to OX in the 

direction OZ, 
y xy = shearing-strain in a plane perpendicular to OX in the 

direction OY, 

and let boc = d>, ^^ = i/r, aob = y, then we shall 

2 2 2 

have 

y Z x = — = tan \f/ , y yz = tan <£, 

y zy = tan <f>, y xz = tan x}/, 

y yx = tan x, y xy = tan \. 

We thus have 

y zy = y yz = tan<£, 
yxz = y 2 ^r = tan if/, 
yxy = y^ = tan x , 

three very important equations. 

We thus have to determine six strains, in order to define 
completely the state of strain in a body at a given point ; viz., 
if we assume three rectangular axes, we must know e x , e y , e z , 
y zy — y yz , yzx = y xz > y xy = ?*» three normal and three tangen- 
tial strains. 



STRAINS IN TERMS OF DISTORTIONS. 



855 



§ 276. Strains in Terms of Distortions. — For the sake of 
clearness, we will consider first only the strains that are parallel 
to the z plane ; hence, will use only two co-ordinate axes, OX 
and OY, as shown in Fig. 297. In this case let us assume a 
small rectangular particle, acbd, the co-ordinates of one corner 
of which are x, y y and of the other, x -f- dx, y -^ dy\ this being 




the case before the load is applied. Let the effect of the load 
be to move the point a to e, and b to /, transforming the 
rectangle acbd into ekfl, and thus changing x, y respectively 
into x -f- £, y + rj, and changing x -\- dx, y -f- dy respectively 
into x + B, -f dx + d%, y -f- 1) + dy + drj. Then are dx, dy 
the sides of the particle before the load is applied. Then 
from what has preceded we shall have 



e^ = 



dZ 

dx 



dr) 



dy 



d£, , dr) 



The first two are evident at once. To prove the third, ob- 



856 



APPLIED MECHANICS. 



serve that the shearing-strain, y xy is the tangent of the angle 
by which the angle kel differs from a right angle ; hence it is 
the tangent of the sum of the angles kem and len. Now, since 
these angles are small, we may take the sum of the tangents 
as nearly equal to the tangent of the sum. But 



Hence 



tan kem = — = — nearly, and 
em ax 



In dH 
tan len = — = -y- nearly. 
en ay J 



d£> , drt 




Fig. 298. 



In the general case, Fig. 298, a rectangular parallelopipedi- 



STRAINS IN TERMS OF DISTORTIONS. 857 

cal particle, the co-ordinates of one corner of which are x, y, z, 
and of the other, x + dx, y -\- dy, z + dz ; this being the case 
before the load is applied. 

Let the effect of the load be to change x, y, z, respec- 
tively, into x -f- £, y + 77, z + C, and to change x -f dx, y -\- dy, 
z + dz/mto (* + S;) + (dx + dS;),(y + v )+(dy + d v ), (? + C) 
+ (dz -f dQ. Then are dx, dy, dz, the edges of the particle 
before the load is applied. 

Then, from what has preceded, we shall have 

_ di d v dt 

dx dy dz 



dH ,d v d£ t% di) , d£ 

Jy + Tx> I--*"****} ^=^ = dz+dy 



The first three will be evident at once. As to the last three, 
the proof is similar to that just used in the case of two co- 
ordinate axes. 

-j + -j- = tan x , — + — = tan^r, -f + — = tan*. 
dy dx dz dx dz dy 



§ 277. Determination of the Strain in any Given Direc- 
tion. — Suppose we are required, knowing the strains e x , c y , e Zj 
Ixyi Ixzi y yz > to determine the strain in a direction making angles 
a, fi, y, with OX, Y, OZ respectively. Assume our rectangu- 
lar parallelopipedical particle in such a way that the diagonal 
from (x, y, z) to (x -\- dx, y + dy, z + dz) shall be in the 
required direction, and call the length of this diagonal ds (Fig. 
298) ; then we shall have 

(dsy = (dxY + (dy)* + (<&)«, (1) 

dx , x 



858 APPLIED MECHANICS. 



cos ^ = |. .. (3) 

dz / . 

cosy = -. (4) 



Let € be the strain in the required direction ; then length a 

diagonal after load is applied will be 

ds{\ + e), 

and we shall have 

(ds) 2 (i + € ) 2 = (dx + d£) 2 + (dy + ^) 2 + (<& + <#)», 
or 
(ds) 2 + 2e(^) 2 + € 2 (^$-) 2 = (dx) 2 + 0^) 2 4- (<&) 2 + 2(dM? 

+ ^ + dzdO 4- (<#) a 4- (drj) 2 + (^) 2 . (5) 

Now, subtracting (1) from (5), and neglecting e 2 (ds) 2 , (d£) 2 , 
(d,j) 2 , and (d£) 2 as being very small compared with the rest, we 
have 

2e(ds) 2 = 2dxd£ + 2dydrj 4- 2dzd£ 



or 
But 



«* = pi + £*, + £* (6) 

dfr ds ds . 



eds = d£cosa + drj cos /3 4- </£cosy. (7) 



* = ^ 4- f ^ + f ^, (8) 

or ^v afe 



STRESSES. 859 

Hence, substituting these, we have, after dividing by ds, and 
observing (2), (3), and (4), 

d£ -> . dn ■> n d£ „ /drj d£\ _ 

€ = —cos 2 a H -'cos 2 /? + — cos 2 y -f ( — + — )cos/?cosy 

dx dy dz \dz dy/ 

/d£ dt\ /dn d£\ ~ , 

+ (s + ^) COSaCOS1, + U + ^)«»««»*' (») 

or, making use of § 276, we have 

< = € x COS 2 a + €y COS 2 /3 + e 2 COS 2 7 -f y yz COS ^8 COS y 

4- y* 2 cosacosy + y^cosacosyS, (12) 

which gives us the strain in any direction. 

It can be shown that there are three directions, at right 
angles to each other, that give the maximum strains or mini- 
mum strains : and we might deduce the ellipsoid of strains, in 
which semi-diameters of the ellipsoid represent the strains ; but 
we will pass on to the consideration of the stresses. 

§ 278. Stresses. — When a body is subjected to the action 
of external forces, if we imagine a plane section dividing the 
body into two parts, the force with which one part of the body 
acts upon the other at this plane is called the stress on the 
plane ; and, in order to know it completely, we must know its 
distribution and its direction at each point of the plane. If we 
consider a small area in this plane, including the point O, and 
represent the stress on this area by/, whereas the area itself is 

P 

represented by a, then will the limit of — , as a approaches zero, 

be. the intensity of the stress on the plane under consideration 
at the point O. Observe that we cannot speak of the stress at 
a certain point of a body unless we refer it to a certain plane 
of action : thus, if a body be in a state of strain, we do not 
attempt to analyze all the molecular forces with which any one 



86o 



APPLIED MECHANICS. 



particle is acted on by its neighbors : but, when we assume a 
certain plane of section through the point, the stress on this 
plane at the point becomes recognizable in magnitude and 
direction ; and what the magnitude and direction of the stress 
at the given point is, depends upon the direction of the plane 
section chosen, the magnitude and direction differing for differ- 
ent plane sections through the point. 

§279. Simple Stress. — A simple stress is merely a pull 
or a thrust. Assume a prismatic body, with sides parallel to 
OX, subjected to a pull in the direction of its 
length ; the magnitude of the pull being P. As- 
sume first a plane section AA normal to the 
direction of P, and let area of AA be A. Then, 
if p x represent the intensity of stress at any 
point of this plane, 




_ P 

P*~ A 

This, which is the intensity of the stress as dis- 
tributed over a plane normal to its direction, may 
! be called its normal intensity. 

fig. 2 99 . Q n t k e other hand, if we desire to ascertain 

the intensity of the stress on the oblique plane BB, making an 
angle with AA, we shall have 



Area BB = 



COS0 



Hence, if p r represent the intensity of the stress on this plane 
in the direction OX, we shall have 



P P 

— : = — COS = p x COS 6. 



(I) 



^cos 



If we resolve this into two components, acting respectively nor 






COMPOUND STRESS. 



86 r 



mal and tangential to BB, and if we denote the normal intensity 
by/«, and the tangential by p t , we shall have 

p n = p r cosO = p x cos 2 6, (2) 

pt = p r sin 6 — p x cos 6 sin 6. (3) 

If, now, we assume another oblique plane section, perpen- 
dicular to the first, we shall obtain the normal //and the tan- 
gential // stress on this plane by substituting for 0, ; 

• 2 

hence we obtain 

pn = p x sm 2 6, (4) 

pt = p x cos0sin6. (5) 



Hence follows 



= 4 



r, 



or, the tangential components of a simple stress on a pair of 
planes at right angles to each other are equal. 

§ 280. Compound Stress. — A compound stress may be 
accounted to be the resultant of a set of simple stresses, and 
may be analyzed into different groups of simple stresses. 

Proposition. — Whatever be the external forces applied to a 
body, if through any point we pass three planes of section at right 
angles to each other, the tangential components of the stress on 
any two of these planes in directions parallel to the third must 
be of equal intensity. 

To prove this proposition, assume 
three rectangular axes, origin at O, and 
assume a rectangular parallelopipedical 
particle, as shown in the figure, so 
small that we may without appreciable 
error assume the stress on any one of 
the faces to be the same as that on the 
opposite face ; resolve these stresses, 
i.e., the forces exerted upon the faces of the particle by the 
other parts of the body, into components parallel to the axes. 



4r 



x,q 



^ 



& 



d? 



% 



Fig. 300. 



862 APPLIED MECHANICS. 

Let <t x = intensity of normal stress on the x plane, 
a-y = intensity of normal stress on the y plane, 
<r z = intensity of normal stress on the z plane, 
r xy = intensity of shearing-stress on x plane in direction 

OY, 
r xz = intensity of shearing-stress on x plane in direction 

OZ, 
Tyx ■=. intensity of shearing-stress on y plane in direction 

OX, 
T yz = intensity of shearing-stress on y plane in direction 

OZ, 
r zx = intensity of shearing-stress on z plane in direction 

OX, 
Tzy — intensity of shearing-stress on z plane in direction 
OY. " 
We have thus apparently nine stresses, which must be given, 
in order to define the stress at the point O completely ; but we 
will now proceed to prove that 

T X y := Ty X , T xz = T ZX , T Z y = Ty Z . 

In the figure, the only ones of these stresses that are repre- 
sented are the following : — 

Xa = x x a x = <r x> 

y yp = yfii = o>, 

yP 2 = yi p 3 = TyXi 
Zy — z 1 y I — cr z . 

The other four are omitted, in order not to complicate the 
figure. 

Now, it is evident that the total normal force on the face 
AFGD and the normal force on the face OBEC balance each 
other independently, and t likewise with the other normal forces. 



GENERAL REMARKS. 863 



The only forces tending to cause rotation around OZ are 
the equal and opposite parallel forces r xy (area AFGD), one act- 
ing on the face AFGD, and the other on the face OBEC ; and 
the equal and opposite forces r yx (area FBEG), one acting on 
the face FBEG, and the other on the face CO AD. 

The first pair forms a couple whose moment is r xy (area 
AFGD) (xx x ), and the second has the moment r yx (area FBEG) 

But 

Area AFGD = (FA) (zz,), area FBEG = (FB) (zz z ) 

.-. r xy (FA) (zz,) (xxj) = r yx (FB) (zz,) (yy x ). 

Cancelling zz» we have 

r xy (FA)(xx 1 ) = r yx (FB)( yyi ). 
But 

FA = yy,. and FB = xx x 

.\ r xy (xx,) (yy x ) = r yx (xx,) (yy z ) 

T xy == T yxy 

Q. E. D. 

In a similar manner we can prove 

T xz == T zx> 



yz — • zy 



GENERAL REMARKS. 



From what precedes, it follows, that, when we have the six 
stresses 

&xi &yi ^Z) T xy> T xz) r yzy 

or, in other words, the normal and tangential components of 
the stresses on three planes at right angles to each other, given, 
the state of stress at that point is entirely determined ; and,. 
when these are given, it is possible to determine the direction 
and intensity of the stress on any given plane. 



864 APPLIED MECHANICS. 



Moreover, if three rectangular axes, OX, OY, and OZ, be 
assumed, and the direct strains along these axes be given, and 
also the shearing-strain about these axes, then the direct strain 
in any given direction can be determined, and also the shearing- 
strain around this direction as an axis. 

The two above-stated propositions furnish two of the funda- 
mental propositions of the theory of elasticity, the third being 
the determination of the relation between the stresses and the 
strains. 

§ 281. Relations Governing the Variation of the Stresses 
at Different Points of a Body. — If we assume a point whose 
co-ordinates are (x, y, z), and a small parallelopipedical particle 
having this point and the point (x -j- dx, y -\- dy, z + dz) for the 
extremities of its diagonal, we shall have, for the edges of this 
particle, dx, dy, dz, respectively. 

Now let the stresses at (x, y } z) be 

Gxy °"yj °"z> T xyt T xz> T yz 7 

i.e, <r x denotes the normal stress on a plane perpendicular to 
OX, and passing through the point (x, y, z), etc. Then, for 
the planes passing through (x -f- dx, y + dy, z -f- dz), we shall 
have the stresses 

v x + d(T x , o-y + da-y, <t z + dcr z , r xy + dr xy , r xz -f- dr XZi r yz + dr yz . 

We may also have outside forces acting upon the particle in 
question : if such is the case, let the components of the result- 
ant external force along the axes be respectively 

Xdxdydz, Ydxdydz, Zdxdydz. 

Now impose the conditions of equilibrium between all the 
forces acting on the particle. To do this, place equal to zero 
the algebraic sum of all the forces parallel to each of the axes 



RELATIONS BETWEEN STRESSES AND STRAINS. 86$ 

respectively, the moment equations having already been incor- 
porated in our demonstration that 

T xy = T yxj r xz == r zxt Tyz == Tzy 

Hence we have three conditions of equilibrium, as follows : — 

( Ox + do x — o x ) dydz + {r xy + dr xy — r xy )dxdz + [t X z + dr xz — r xz )dxdy + Xdxdydz = o, 
{o y +do y — a y )dxdz + {Txy+dTxy — T X y)dydz +{Ty Z +d~y Z — T yz )dydx+ Yd x dydz = o, 
(o z +do z —o z )dxdy + {Ty Z +dTy Z —r yz )dxdz J r{Txz^-dr X z — T X z)dzdy + Zdxdydz = O. 

Hence, reducing, and dividing by dxdydz, we have 

&> + *a + ^ + x = (I) 

d& ^ ^ 

^+^+^ + K=o, (2) 

ax ay dz 

^£ + ^ + ^ +Z=0 . (3) 

dx dy dz 

If the particle is in the interior of the body, and we dis- 
regard its weight, then X = Y = Z = o. 

Equations (i), (2), and (3) give the necessary relations which 
the variations of stress from point to point must satisfy in order 
that the conditions of equilibrium may be fulfilled. 

§ 282. Relations between the Stresses and Strains. — 
Before proceeding to the general problems of composition of 
stresses, i.e., of determining from a sufficient number of data 
the stress upon any plane, we will first discuss the relations 
between the stresses and the strains ; and we will confine our- 
selves to those bodies that are homogeneous, and of the same 
elasticity throughout. 

From what we have already seen, if to a straight rod whose 
cross-section is A there be applied a pull P in the direction of 



866 APPLIED MECHANICS. 

its length, the intensity of the stress on the cross-section wi 
be 

P 
° = a'> 

and, if E be the tensile modulus of elasticity of the material of 
the rod, the strain in a direction at right angles to the cross- 
section, or, in other words, in the direction of the pull, will be 



Now, another fact, which we have thus far taken no account 
of, is, that although there is no stress in a direction at right 
angles to the pull, or, in other words, although a section at 
right angles to the above-stated cross-section will have no stress 
upon it, yet there will be a strain in all directions at right angles 
to the direction of the pull : and this strain will be, for any direc- 
tion at right angles to the pull, 



being of the opposite kind from « ; thus, if € is extension, € x is 
compression, and vice versa. 

Hence, if, at any point O of such a rod, we assume three 
rectangular axes, of which OX is in the direction of the pull, 
and we use the notation already adopted, we shall have 

P 

<r x = — , a y = cr z =■ r xy = r xz = r yz = O, 



I <r * e 



X 



E m E m 



fxy = 7xz = Jyz = O. 



RELATIONS BETWEEN STRESSES AND STRAINS. 867 
MODULUS OF SHEARING ELASTICITY. 

In the case of direct tension or compression, when only a 
simple stress is applied, we have defined the modulus of elas- 
ticity as the ratio of the stress to the strain in its own direction. 

Adopting a similar definition in the case of shearing, we 
shall have 

T xy T xz T yz ^, 

yxy Jxz Jyz ~ 

where G is the modulus of shearing elasticity. 



GENERAL RELATIONS BETWEEN STRESSES AND STRAINS. 

Whenever a compound stress acts on a body at a given 
point, let the stresses be 

O'xy <Ty } (T Z} Txyy T XZt Tyz y 



then we shall have, for the strains. 



c, 



0£ _ J_o> _ ]_<?z __ Txy 

E mE tn£ lxy "'' G' 



<Ty I <T X I <T Z _ T xz 

€ y — i- 7-. 7-.' Ixz — ~77> 

E m E m E G 



<J Z I <T X I (Ty Ty Z 

^"E'mE'mE' lyz " G' 



This enables us to determine the strains in terms of the 
stresses, as soon as the values of E, G, and m are known from 
experiment, for the material under consideration. 

If, on the other hand, the stresses be required in terms of 
the strains, we can consider * x , e yi e z , y xy , y xz , y yz , as known, and 
determine <r x , o- y , <t z , r xy> r yzi r x „ from the above equations. 



"\ 



APPLIED MECHANICS. 



We thus obtain 

<Ty + <T Z 

Ei x = <r x — » (i) 

Ety = 0> , (2) 

°\j: "f" °"y 
^ = o-, ', (3) 

and, by solving these equations for the stresses, we have 



w 



m 



m + 



1 \ m — 2 / 



m J ** + *? + e A 

a» + 1 \ m — 2 J 



(5) 
(6) 



and also 

r xy = 67.^, (7) t^ = Gy xz , (8) r^ = Gy yz . (9) 

These equations express the stresses in terms of the strains. 
The three last might be written as follows (see § 276) : — 



<xy 



lyZ 

as these forms are often convenient. 

§ 283. Case when 0-3 = 0. — Inasmuch as there are many 
cases in practice where the stress is all parallel to one plane, 
and where, consequently, the stress on any plane parallel to 
this plane has no normal component, it will be convenient to 
have the reduced forms of equations (4), (5), and (6) which 
apply in this case. 



= G \dy + i) 


(10) 


\dz dx) 


(11) 


c( d " + dl \ 


(12) 



VALUES OF E, G, AND m 869 

Let the plane to which the stresses are parallel be the Z 
plane ; then <r z = o. Then equation (6) becomes 

, *x + €y + e 2 
€ H = O 

m — 2 

. - _ e ^ + <* . 

• . c z — — 



(w - 1) ' 



and, substituting this value of e 2 in (4) and (5), and reducing, we 
obtain 



m 



. E L + 5l±vV (i) 

i\ m — \f v ' 



' m + 1 \* m — 1/ * ' 



which are the required forms. 

The other three equations, viz., — 

r xy = 67,^, T xz = 67^, Tya = Gjyst 

remain the same as before. 

§284. Values of E, G, and m. — These three constants 
need to be known, to use the relations developed above. 

i°. As to E, this is the modulus of elasticity for tension, 
and has been determined experimentally for the various mate- 
rials, as has been already explained. Moreover, it has also been 
shown experimentally, that, with moderate loads, the modulus 
of elasticity for compression is nearly identical with that for 
tension in cast-iron, wrought-iron, and steel. 

2°. As to m, in those few applications that Professor Ran- 
kine gives of his theory of internal stress, such as the case of 
combined twisting and bending, he determines the greatest in- 
tensity of the stress acting ; and his criterion is, that this shall 
be kept within the working-strength of the material. This is 
equivalent to assuming m — 00. The more modern writers, 



870 APPLIED MECHANICS. 

such as Grashof and others, take account of the fact that m has 
a finite value, and make their criterion that the greatest strain 
shall be kept within the quotient obtained by dividing the work- 
ing-strength by the modulus of elasticity of the material. 

Thus, if f is the working-strength, and a- I the greatest 
stress, and c, the greatest strain, Rankine's criterion of safety 
is 

whereas the more modern criterion is 

The resulting formulae differ in each case ; and, as has been 
stated, those of Rankine could be derived from the more gen- 
eral ones by making 

1 

— = o or m = oo, 

which is never the case. 

As to the value of m> but few experiments have been made. 
Those of Wertheim give, for brass, 2.94 ; for wrought-iron, 3.64. 

The values m = 3 and m = 4 are those most commonly 
adopted, so that 

11 11 

— — - or ~ — -• 
m 3 m 4 

3 . The value of G, the shearing-modulus of elasticity, i.e., 
the ratio of the stress to the strain for shearing, has been 
determined experimentally, and has generally been found to be 
about two-fifths that for tension. 

According to the theory of elasticity, we must have 

G = i -2—B, 

2 *n + 1 



as may be proved as follows : — 



VALUES OF E, G, AND m. 87 1 

Assume a square particle whose side is a, and let a simple 
normal stress o- be applied at the face AB ;• then we 
shall have, on the planes BD and AC, a shearing-stress 

(§ 279) 

t = o- sin 45° cos 45° = -J<r. 

On the other hand, if we let 



d 

B 




d 

€ = TT* Fig. 301. 

the strain of the particle in the direction AD will be e, while 
that in the direction AB will be ; hence the particle will 



become a rectangle, the side AD changing its length from a to 

a* 

a -f- at, and side AB changing from a to a . 

wi 

The diagonals will no longer be at right angles to each 

other ; and, if we denote by a the angle by which their angle 

differs from a right angle, we shall have, for the shearing-strain 

on the planes A C and BD, 

y = tan a. 

But, after the distortion, the angle ADB will become 



£-) 



a ae € 

i — tan- a 1 

m m m 

tanf ] = = — : = — r— ; 

a* 1 i= c 



(ir a\ _ 2 _ 

42/ . . a ~~ a 4- 

^ ' 1 + tan - 



therefore, dividing, and carrying the division only to terms of 
the first degree, we have 






m + 1 
= — —* 



872 APPLIED MECHANICS. 



But 



*y = tan a = 2 tan - nearly 
2 

Z8 -f- I 



m 



— ^°" _ 1 m /<r\, 

" W + I ~ 2W + I\€/ 



but 



I = £ and - = E 
7 € 

- r m -E. 



2 m -\- 1 



§ 285. Conjugate Stresses. — If the stress on a given plane 
at a given point of a body be in a given direction, the stress at 
the same point on a plane parallel to that direction will be 
parallel to the given plane. Let YO Y represent, in section, a 
given plane, and let the stress on that plane be in the direction 

xox. 

Consider a small prism ABCD within a body, the sides of 
whose base are parallel respectively to XOX and YOY. The 
forces on the plane AB are counterbalanced by the forces 
on the plane DC; the resultants of each of these sets being 
equal and opposite, and acting along a line passing through O. 
Hence the forces acting on the planes AD and BC must be bal- 
anced entirely independently of any of the forces on AB or 
DC: and this can be the case only when their direction is paral- 
lel to YOY; for otherwise their resultants, though equal in 
magnitude and opposite in direction, would not be directly 
opposite, but would form a couple, and, as there is no equal and 
opposite couple furnished by the forces on the other faces, equi- 
librium could not exist under this supposition. 

§286. Composition of Stresses. — The general problem of 
the composition of stresses may be stated as follows : — 






PROBLEM. 



873 



Knowing the stresses at a given point of a strained body 
on three planes passing through that point, to find the stress at 
the same point on any other plane, also passing through the 
same point. The stresses on the three given planes are not 
entirely independent ; in other words, we could not give the 
stresses on these three planes, in magnitude and direction, at 
random, and expect to find the problem a possible one. Thus, 
suppose that the planes are at right angles to each other, we 
have already seen that we have the right to give their three 

• zf and the three tangential, 
etc. We will now proceed 



normal components, <r x , <r y , and 
T X yi Tyz, and r XZi and that r yx = r, 



xyi 



to special cases. 

§ 287. Problem. — Given the three planes of action of the 
stress as the x, y, and z plane respectively, and given the nor- 
mal and tangential components of the stresses on these planes, 
viz., a- xf a-y, a z , r xyy t xz , and r yz , to find the intensity and direction 
of the stress on a plane whose normal makes with OX, OY, 
and OZ the angles a, ft and y respectively, where, of course, 

COS 2 a + COS 2 /? + C0S 2 y = I. 

Draw the line ON, making angles a, /?, and y with OX, O Y, 
and OZ respectively ; then draw 
near the plane ABC perpen- 
dicular to ON. It has the direc- 
tion of the required plane, and 
cuts off intercepts OA, OB, and 
OC on the axes ; and, moreover, 
we shall have, from trigonometry, 
the relations, 

Area BOC = (ABC) cos a, 

Area AOC = (ABC) cos ft 

Area A OB = (ABC) cosy. 




Fig. 3° 2 ' 



Now consider the conditions of equilibrium of the tetrahe- 
dron OABC. The stress on ABC must be equal and directly 



874 APPLIED MECHANICS. 

opposed to the resultant of the stresses on the three faces 
AOC, BOCy and A OB. Now let us proceed to find this result- 
ant. 

In the direction OX we have the force 

<r x (BOC) + r xz (AOB) + r xy (AOC) 

= (ABC)(a x COS a + T X yCOsP -f T* z COSy). 

Lay off OD to represent this quantity. In the same way repre- 
sent the force in the direction O Y by 

OE = <r y (AOC) + r yz {BOA) + r xy (BOC) 

= (ABC) (cr,, cos /3 -}- T^cosa + T^cosy), 

and that in the direction OZ by 

OF = <t z (AOB) 4- r xz (BOC) + r„(AOC) 

= ABC(cr z cosy + t** COS a + r yz C0s/3). 

Now compound these three forces, and we have, as resultant 

force, 

R = OG = S/OB 2 + <2£> + tfi? 2 , 

and as resultant intensity 



^ V6>Z> 2 + OE 2 -f OF 2 



= V^(°"-* C0Sa + T xy cos f3 4- r^cosy) 2 

+ ((T y cos/3 -\- r X y COS a 4- r^COSy)* 

4- (a* cos y 4- t^ cos a 4- r^ cos /3) 2 \ 

= ^\<T x 2 COS 2 a 4- a/ COS 2 /? + cr 2 2 C0S 2 y 

4- T^, 2 (cos 2 a 4- cos 2 /?) + Tr 2 2 (cos 2 a 4- cps 2 y) 

4- Tj, z 2 (cOS 2 /? + C0S 2 y) 4- 2 (T x (T xy COS ft 4" T^ COS y) COS a 

4" 2a-y(T xy COSa. 4" T yz COS y) COS ft 
4- 2 a* (t x * COS a 4- Ty Z COS ft) cosy + 2T^T* a COS /? cos y 

4- 2r X yTyz COS a COS y 4- 2 Ty Z T xz COS a COS /?| ; 



STRESSES PARALLEL TO A PLANE. 



875 



the direction being given by the angles, a r , $ r , and y r , where 

COS a r 



OP 
R ' 



a OE OF 

cos/? r = — , cosy r = — . 



§ 288. Stresses Parallel to a Plane. — To solve the same 
problem when there is no stress in the direction OZ, and when 
the new plane is perpendicular to XOY, or, in other words, in 
the case when the planes of action are all perpendicular to one 
plane, to which the stresses are all parallel :' we then have 



o* 



t xz = Tyz — o and (3 = 90 — a, 



and hence 

a- = Vo\r 2 COS 2 a + ay* sin 2 a + r'jgf + 2(a x -f- cr y ) r xy COS a sin a. 

Or we may proceed as follows : — 

Let the normal intensity of the stress on the x plane (i.e., 
that perpendicular to OX) be <r x , that on the 
y plane a yy and the tangential intensity r xy . 
Let ON be the direction of the normal to 
the plane on which the stress is to be deter- 
mined, and let the angle XON = a. Then 
let the plane AB be drawn perpendicular to 
ON, and let us consider the equilibrium of 
the forces exerted by the other parts of the 
body upon the triangular prism whose base is ABO and alti- 
tude unity. 

If we compound the forces acting on the faces AO and OB, 
we shall have, in their resultant, the total force on the face AB 
in magnitude and direction. Moreover, we have the relations, 

Area OB = area AB cos a and Area OA = area AB sin a. 
Force acting on OB in direction OX = cr x ( OB) , 
Force acting on OB in direction OY = Tj 
Force acting on OA in direction OX 
Force acting on OA in direction OY = cr y (OA), 




Fig. 303. 



*y(OB), 

Scy^OA), 



876 APPLIED MECHANICS. 

Hence, if we lay off 

OD = <r x (OB) + T xy (OA) and OC = <r y (OA) + r xy {OB), 

then will <9Z> represent the total force acting in the direction 
OX, and OC will represent the total force acting in the direc- 
tion OY. 

Compounding these, we shall have OE as the resultant total 

OF 
force on the face AB, and — — witt represent its intensity. 

AIj 

To deduce the analytical values, we have 
OD = <r x (OB) 4- T xy (OA) = (^9) (oleosa 4- r^sina), 

OC = <Ty(OA) + T xy {OB) = {AB){<Ty SHI a + T^ COS Cl) 

.\ OE = \/OD 2 + <9C 2 



= AB\ (<r x COS a -f- Tjpy sin a) 2 -f ((Ty sin a 4- t^ cos a) 2 

= AB\ \ <t x 2 cos 2 a + Oj, 2 sin 2 a + 2T^y cos a sin a(ar x + o>) 

4- T jr /(cos 2 a 4- sin 2 a)£. 

Or, if oy represent the resultant intensity on the plane AB, and 
a^ the angle this resultant makes with OX, we shall have 

o> = \\a- x 2 cos 2 a 4- oy sin 2 a 

4- 2T X y(o- x 4- o>) COS a sin a 4- T>/j, (i) 



and 



OD , . <9C 

costv = and sin<v = — . 

OE OE 



Moreover, it is sometimes desirable to resolve the stress into 
normal and tangential components. If this be done, and if a 
and r represent respectively the normal and tangential com- 
ponents, we shall have 

OF , EF 

cr = — — and t = -— : 
AB AB' 



PRINCIPAL STRESSES. %jy 

but 

OF = Oleosa + EDsma and EF = BZ>cosa — OZ>sina 

on . oc . 

cr = COS a + Sin a 

AB AB 

= cr^COS 2 a + cr^ sin 2 a + 2t^ COS a sin a (2) 

and 

OC OD . 

t = COS a — Sin a 

AB AB 

= (<Ty — o-.*) cos a sin a + T^(cos 2 a — sin 2 a) 

(<Ty <T X \ 
J sin 2a 4- r xy cos 2a. (3) 

§ 289. Principal Stresses. — It will next be shown, that r 
whatever be the state of stress in a body, provided the stresses- 
are all parallel to one plane, the planes of action being all taken 
perpendicular to this plane, there are always two planes, at right 
angles to each other, on which there is no tangential stress ; 
these two planes being called the planes of principal stress, the 
stress on one of these planes being greater, and the other less, 
than that on any other plane through the same point. 

To prove the above, it will be necessary only in the last 
case, which is a perfectly general one, to determine for what 
values of a the value of t is zer.o, and whether these values of a, 
are always possible. We have 

°"v — °"x 

r = sin 2a -j- t xv cos 2a : 

2 xy 

and, if we put this equal to zero, we have 



Sin 2a 2Tj 

= tan 2a = 



->' 



and this gives us, for all values of cr x , <r yy and r xy , two possible 
values for 2a, differing from each other by 180 , hence two 
values for a differing by 90 . Hence follows the first part of 
the proposition. 



'8; 8 APPLIED MECHANICS. 



The latter part —that these are the planes of the greatest 
and least stresses —will be shown by differentiating the; value 
of o> 2 , and putting the first differential co-efficient equal to zero ; 
and, as this gives us 

— 2<ix cos a sin a -f- 2cr/ cos a sin a 
+ 2T xy ((T x + Oj,)(cos 2 a — sin 2 a) 

= 2 (cr x -f- (T y ) \ (a-y — cr x ) COS a sin a -f TOCOS' a — sin 2 a)\ 

= o, 

therefore we have the same condition for the maximum and 
minimum stresses as we have for the planes of no tangential 
stress. 

It follows that the determination of the greatest and least 
stresses at any one point of a body is identical with the deter- 
mination of the principal stresses ; and it will be necessary, 
whenever the stresses on any two planes are given, to be able 
to determine the principal stresses, as one of these is the 
greatest stress at that point of the body, and the other the 
least. 

§ 290. Determination of Principal Stresses. — When the 
stress is all parallel to one plane, viz., the z plane, and when 
the stresses on two planes at right angles to each other are 
given, i.e., their normal and tangential components, we may be 
required to determine the principal stresses. Proceed as fol- 
lows : Given normal stresses on X and Y planes respectively, 
<t x and ay, and tangential stress on each plane r xy} to find prin- 
cipal stresses. 

From § 288 we have, for a plane whose normal makes an 
angle a with OX, 

.OV = yo-^2 cos 2 a _|_ o-y s i n 2 a _j_ 2T xy (a x + o-y) cos a sin a + r Xy 2 , (1) 

a- n = o\*COS 2 a + CTj,sin 2 a -|- 2r xy cos a sin a, (2) 

r = T^,(C0S 2 a — sin 2 a) — (or x — a-y) COS a sin a, (3) 



DETERMINATION OF PRINCIPAL STRESSES. 879 



or 



xy 



(4) 



Now, the condition that the plane shall be a plane of prin- 
cipal stress is, that r = o. Hence write 

Try (COS 2 a — sin 2 a) — (cr x — oy) COS a sin a = O, 

find a, and substitute its value in (2), and we shall have the 
principal stresses. The operation may be performed as fol- 
lows ; viz., — 

From (3) we have 

<?x — <?y 



(a) 2 cos 2 a — 1 = — cos a sin a 

1 ( , <T* ~ <Ty . I 
a = -<H — COS a Sin a > 

2 ( T XV ) 



xy 

.*. COS 2 



'x — u y 



{&) 1—2 sin 2 a = COS a sin a 

Txy 



sin 

^xy 

Hence 

cr* + a-y COS a sin a j cr x 2 — 2cr x <Ty + o* : 2 



1 J or, — a> \ 

a = - < 1 COS asina . 

2 ( r^ } 

j OV 5 — 2(T x <Ty + O-/ j 

i ^ + 4T *4 



or 



<*x + ^v cos a sin a ( , 

<7„ = 1 + . \(cr x - o>) 2 + 4 t^ 2 J, 



But we have, since (4) equals zero, 



2T xy 

tan 2a = 



0* — °> 



sin 2a = 2 sin a cos a = 



v^a* - <r y y + 4^/" 



88o APPLIED MECHANICS. 

Hence substitute for cos a sin a its value, and 

Vn = ^±^> ± jV(ct,-o > )»+4t^", (5) 

which gives us the magnitudes of the principal stresses ; the 
plus sign corresponding to the greater, and the minus sign to 
the less. 

EXAMPLES. 

i. Let, in the last section, <r y = o, and find the principal stresses. 
Here we have 

2T X y 

tan 2a = 

and 

<T X I I 

<?n = — ± -\<T X 2 + 4T*/. 

2 2 

2. Given two principal stresses, to find the stress on a plane whose 
normal makes an angle a with OX. 

In this case r xy = o. 

Hence we have the case of § 288, with the reduction of making 
r xy = o. We may therefore obtain the result by substitution in the 
results of § 288, or we may proceed as follows : — 

(a) Find stress on new plane in direction OX; this will be, § 279, 

<t x cos a. 

(b) Find stress on new plane in direction OY; this will be, § 279, 

a-y sin a. 

(c) Compound the two, and the resultant is 



o> = Vo-* 2 COS 2 a + a-y 2 sin 2 a. (i} 

(</) Normal component of <r x cos a is 

<r x COS 2 a. 

(f) Normal component of <r y sin a is 

<Ty sin 2 a. 



ELLIPSE OF STRESS. 



8&i 



(/) Add, and we have, for normal stress, 

o-„ = <t x cos 2 a -f- o> sin 2 a. 
(£•) Tangential component of a x cos a is 
— a x cos a sin a. 
{h) Tangential component of a y sin a is 
-f-a^ cos a sin a. 
(k) Add, and we have, for tangential stress, 
t = (o-j, — o-.*) cos a sin a. 



(*) 



(3) 



§291. Ellipse of Stress In the case above, i.e., when 

the two principal stresses are a x and a y respectively, if we 

represent them graphically by OA = 

a x and OB = a- y} and let CD be the 

plane on which the stress is required, 

its normal making with OX the angle 

XON = a, then, from what has been 

shown, if OR represent the intensity of 

the resultant stress on this plane, we 

shall have 



OR = <r r = Vo-^ 2 cos 2 a + oy 5 sin 2 a/ 
and, moreover, 




OE = oleosa, 



OR = <r y sin a. 



If we denote these by x and y respectively, letting (x, y) be 
the point R, i.e., the extremity of the line representing the 
stress on AB, then 



X = ovCOSa, 



ay Sin a, 



•\ ( — J = cos 2 a and ( — J = sin 3 






882 APPLIED MECHANICS. 

which is the equation of an ellipse whose semi-axes are <t x and 
a-y respectively ; hence the stress on any plane will be repre- 
sented by some semi-diameter of the ellipse. 



SPECIAL CASES. 

I. When the two given stresses are equal, or cr x = oy, then 



v r = Vo* 2 COS 2 a 4- o-y 2 sin 2 a = o>, 

and 

a x cos a . 

costv = = cos a and sm<v = sin a/ 

therefore the stress is of the same intensity on all planes, and 
always normal to the plane. 

II. When the two given stresses are equal in magnitude 
but opposite in sign, or <r y = —a- xf then 



But 

cos<v = cos a and sina r = —sin a, 
hence 

Or = —a; 

therefore the stress on any plane whose normal makes an angle 
a with OX is of the same intensity o- xy but makes an angle 
equal to a with OX on the side opposite to that of the normal 
to the plane. 

Problem. — A pair of principal stresses being given, to 
find the positions of the planes on which the shear is greatest. 

Solution. — Let t = (o- y — o- x ) sin a cos a' = max. 

Therefore differentiate, and 

cos 2 a — sin 2 a = o 
•\ cos a = ±sina .\ a = 45 or 135 . 



SPECIAL MODES CS SOLUTION OF SOME PROBLEMS. 883 

§ 292. Some Special Modes of Solution of some Prob- 
lems. — The case where two principal stresses, a x and a y , are 
given, to find the stress on any plane whose normal makes an 
angle a with OX, may be solved as follows, graphically : — 

Let, Fig. 304, <t x — OA, and a y = OB. Let XON = a. 

Now, 



°> 


= 


°> 


+ 
2 


<T* 


+ 


°> 


2 


0\* 

> 


<T .. 




°> 


+ 


°* 




°> 


— 


O-^r 



Hence, instead of proceeding at once to find the resultant 
stress on CD due to the action of a x and a y , we may first find 
that due to the action of the two equal principal stresses of the 
same kind, 

q> + q* 
2 ' 
then that due to the pair 

o> — <r* , a y — a x 

— and , 



and then the resultant of these two resultants. 

The first resultant will be evidently laid off on ON, and 

equal in magnitude to — ; hence let OM = — -, and 

OM will be the first resultant. 

The second resultant will be of magnitude — -, and 

will have a direction MR such that the angle NMS = SMR. 

Hence, laying off this angle, and making MR = — -, 

we shall have for the final resultant, OR, as before. 

This construction will be useful in the following case: — 
To find the most oblique stress, we must find for what 

value of a the angle MOR is greatest. This will be made 



884 APPLIED MECHANICS. 

evident if we observe, that, for all positions of the plane, the 
triangle OMR has always OM — — — , and MR = " 



2 2 

both of constant length. Hence, if, with M as a centre and 
MR as a radius, a circle were described, and a tangent were 
drawn from to this circle, the point of tangency being taken 
for R, then will OR be the most oblique stress ; i.e., the 
stress is most oblique when ORM = 90 . Therefore greatest 
obliquity = 



sin- 



. T °> ~ ^ 
o> 4- o-j 



§ 293. Converse of the Ellipse of Stress. — The converse 
of the ellipse of stress would be the following problem : Given 
any two planes passing through the point in question ; given 
the intensities and directions of the stresses on these planes, 
— to find the principal stresses in magnitude and in direction. 

The first step to be taken is, to assure ourselves that the 
conditions are not incompatible, as they are liable to be if the 
planes and stresses are taken at random. The test of this 
question is, to resolve each stress into two components, respec- 
tively parallel to the two planes ; and, if the conditions are not 
inconsistent, the components of each stress along the plane on 
which it acts must be equal. The proof of this statement can 
be made in a similar way to that used in proving that the 
intensities of the shearing-stresses on two planes at right angles 
to each other are equal. If, upon applying this test, we find 
that the conditions are not inconsistent, we may proceed as 
follows : — 

Suppose CD (Fig. 304) were the given plane, and OR the 
stress upon it, and suppose the position of the principal axes, 
OX and O Y, and, indeed, all the rest of the figure, were absent, 
i.e., not known. Now, we can easily draw the normal ON; 
and, if we could determine upon it the point M such that OM 



CONVERSE OF THE ELLIPSE OF STRESS. 885 

should be one-half the sum of the principal stresses, we should 
be able to reproduce the whole figure. Hence we will devote 
ourselves to the determination of the position of the point M. 

Let OR = p = stress on plane CD. 

Let stress on the other given plane be/ x . 

Let NOR — = obliquity of /. 

Let 0, = obliquity of /,. 

Then we have 

MR 2 = OR 2 + OM 2 - 2OM. ORcosO; 

or, if <t x and <r y denote the (unknown) magnitudes of the prin- 
cipal stresses, 

/o> - CT X \ 2 f(T x + (T y \ 2 (<T X + o>\ 

From the triangle constructed in the same way, with the 
stress on the other plane, we should have 

(<T y — <T X \ (<T X + 0>\ 2 (<T X + <Ty\ ■ 

\—T-J =■* + V~T~ ) -*(—;— J* cos fc. (2) 

Hence, by subtraction, 

p*-'p* = J ^ x + * y J {p cos 0-p s cos 6 X ) (3) 

<Tx + °> P 2 — Px 



2 (/ cos — p t cos 6 l ) 



(4) 



o",+ 



Having thus found , we can next find, from either 



1) or (2), the value of — 



<Tv — cr. 



2 

Now, therefore, we know OM and Vkfi?, and hence we can lay 
off this value of OM, and complete the triangle OMR ; then 



S86 APPLIED MECHANICS. 




bisect the angle NMR, and the line MS is parallel to the axis 
of greater principal stress. Hence draw Y parallel to MS, 
and OX perpendicular to OY, and lay off on OY 

0B= a v = 0M+ MR, 
and on OX 

0A = <r x = OM—MR, 

and the problem is solved. 

§ 294. Rankine's Graphical Solution. — The following is 
Rankine's graphical solution of the preceding problem : 

Draw the straight line ON, Fig. 305, and then lay off angle 
NOR = 0, and angle NOR, 
= 6-, also OR=p and OR, o 
= p 1 . Then join R and R, y 
and bisect RR Z by a perpen- 
dicular SM. From the point 
My where this meets 0N f draw MR and MR, . Then will 



°±+-^=OMy (1) - V —^ = MR = MR X ; (2) 



.'. <r x = 0M- MR, (3) and <r v = OM+MR ; (4) 

and the angles made by Y with OWand ON, , respectively, 
will be 

NMR , . , o „ NMR, 
9 o°-a = — — , (5) and 90 - «» = — - 

A comparison of this figure with the triangle OMR of 
§ 291 will show that this is merely a graphical construction for 



RANKINE'S GRAPHICAL SOLUTION. 



887 



the analytical solution given in § 293 ; and the equations of 
that article can readily be deduced from the figure given above. 



SPECIAL CASES. 



(a) When the two given planes are at right angles to each 
other. — In this case the tangential components of OR and 0R X 
are equal, and hence (Fig. 306) aR x = bR. 

Hence the figure be- 
comes that shown where 
RR Z is parallel to ON; and 
if we let Ob = p n , Oa = p n \ , 
and bR — aR z —p t ,we shall 
have 




SL+^ =tw= A±A_'. 



mr = V{Mby +p t 2 = \/{ ? n /*y +tf 



whence we readily obtain <r x and <r y , and then a and a 1 , just as 
before. 

(b) When the two given stresses are conjugate (see § 285). — In 
this case the obliquities of the two stresses are the same, and 
the figure becomes Fig. 307. 

We then obtain M 



<r x 4- (T y 



= OM 



OS 



cos 



~e 




= *±4; (9) 

2 cos o 



Fig. 



888 APPLIED MECHANICS. 



<r v — (T 



n — ■ = mr= V{sry + {Msy 



= |/(^-=A) 2 + OS* tan' a = */(£=A )' + [l±*) % tan' 



= /(^)V + tan 8 6)-pp i = i/&±gT -pp x . ( ie ) 



(c) The following proposition, due to Rankine, will next 
be proved : 

The stress in every direction being a thrust, and the great- 
est obliquity being given, it is required to find the ratio of two 
conjugate thrusts whose common obliquity is given. 

Let denote the greatest obliquity, then we shall have (see 
last equation of §292) 



^-^ = 8^0. (n) 



Now let 6 denote the common obliquity of two conjugate 
stresses whose intensities are/ and/ x . Moreover, < 0, and 
we will consider p 1 < p. Then from equations (9) and (10) we 
deduce 

4pp, cos 2 _ l (T y — o- x y 
1 0> + A) 2 ~ U. + o-J ; 

and combining this with (n) we obtain 



4pp } cos 3 



(2) 



RANKINGS GRAPHICAL SGLUT10N. 889 



( P ~A V _ COS 3 6 — cos 2 
''' V/+/J ~ C0S 2 # ; 

^, cos — V cos 2 — cos 2 

/ cos + 1/ cos 2 — cos 2 

wnich is the ratio shown. 

§ 295. Case of any Stresses in Space. — In the case of 
stress which is not all parallel to one plane, we should find that 
it is always possible, no matter how complicated the -state of 
stress in a body, to find three planes at right angles to each 
other on which the stress is wholly -normal, these being the 
principal stresses ; and a number of propositions follow analo- 
gous to those for stresses all parallel to one plane. The discus- 
sions of these cases become very complex, and will not be 
treated here. 

§ 296. Some Applications. — The following are some of 
the practical cases which require the theory of elasticity for 
their solution. 

§297. Combined Twisting and Bending. — This is the 
case very generally in shafting, as the twist is necessary for 
the transmission of power, and the bending is due to the weight 
of the pulleys and shafting, and the pull of the belts, this being 
especially so when there are pulleys elsewhere than close to the 
hangers ; also in overhanging shafts, in crank-shafts, etc. 

Thus far we have no tests of shafting under combined 
twisting and bending, and therefore the methods used for 
calculating such shafts vary. With many it is the practice 
to compute their proper size from the twisting-moment only, 
but to make up for the bending by using a large factor of 
safety, the magnitude of this factor depending upon how much 



§9° APPLIED MECHANICS. 

the computer imagines the shaft will be weakened by the par- 
ticular bending to which it is subjected. 

With Others it is customary to compute the deflections, 
under the greatest belt-pulls that can come upon it, by the 
principles of transverse stress, without any reference to the 
torsion, and to so determine it that the deflection computed in 
this way should not exceed -Y2V0 or iqo~o °^ tne s P an - 

On the other hand, Unwin and some others give the for- 
mulae, which will be developed here for combined twisting and 
bending, as deduced by the theory of elasticity. This formula 
has not, as yet, been very extensively used ; and its constants 
are taken from experiments on tension or torsion alone, and 
not on a combination of the two. It is to be hoped that we 
may some time have some experiments on such a combination. 
We will now proceed to deduce a formula for the greatest in- 
tensity of the stress at any point of the shaft. 

For this purpose 

Let M x = bending-moment at any section. 

M 2 = twisting-moment at the same section. 

/, = moment of inertia about neutral axis for bending. 

f 2 = moment of inertia about axis of shaft. 

r = distance from axis to outside fibre. 

Then, if we denote by o- the greatest intensity of the stress 
due to bending, and by t the greatest intensity of the stress due 
to twisting, we have, 

Ms M 2 r , v 

<r = ~f- (l) r = — • (2) 

For a circular or hollow circular shaft, 

/ 2 =2/ i; 



hence 



M x r . v M 2 r / v 

" = X- <3) T= 17T (4) 



COMBINED TWISTING AND BENDING. 89 1 

Then, at a point at the outside of the shaft in the section 
under consideration, we shall have, — 
1°. On a plane normal to the axis, 

(a) a normal stress <r, 

(b) a shearing-stress r. 

2°. On a plane in the direction of the axis, 

(a) a normal stress o. 

(b) a shearing-stress r. 

We thus have the case solved in Example I., § 290. 
If, therefore, the greatest and least principal stresses be 
denoted by o-j and o- 2 respectively, we shall have 






V?+* <»> 



2 '4 



+ r*. (6) 



But, if e, and e 2 denote the strains in the directions of the 

principal stresses, we have 

77 °" 2 c °" 1 . 
£ei = o-j — — > ZS€ 2 = <r 2 j 



Hence, substituting for <r l and o- 2 their values, we have 



1 . m 4- 1 



*,, = =__i, + '^L^^-r^—r, ( 7) 

2m 2M 



m — 1 w + 1 



*% = icr - ^^-ty„. + 4^. W 



2M 



We then have, for the greatest stress on any fibre, the 
greater of the two quantities (7) and (8); and this should not at 
any section of the shaft exceed the working-strength of the 
material for tension. 



892 APPLIED MECHANICS. 

The greater of the two is Ec s : hence we should have, if 
f = greatest stress, 



1 m -f 1 



,+ V? + 4t--/ ( 9 ) 



2#2 2/tf 



If, now, we let m = 4, as is commonly done, we have 



1^ + 1^ + 4^=/, (10) 

this being the formula given by Grashof and others for com- 
bined twisting and bending. 

On the other hand, Rankine puts the value of o- x in (5) equal 
to f> and hence Rankine's formula is 



<T 1 



2 + -VW 4 r*=/. (II) 



This might be derived from (9) by making m = 00 instead 
of m = 4. 

The formulae developed above are applicable to any section. 

APPLICATION TO CIRCULAR AND HOLLOW CIRCULAR SHAFTS. 

Substituting for o- and t in (10) the values from (3). and (4), 
we should obtain 



jfyf x + \\ImFTms\ =/, (12) 



which is Grashof s'formula, and is given by Unwin and others; 
and, substituting in (11) instead, we should have 



-jW + SIM- -f M 2 >) =/. (13) 

Equation (12) is equivalent to the following rule:— 



HOLLOW CYLLNDERS SUBJECTED TO PRESSURE. 893 

Calculate the shaft as though it were subjected to a bending- 
moment 

M = \M, + i\ImF+~W', 

and equation (13) is equivalent to the following rule : — 

Calculate the shaft as though it were subjected to a bending- 
moment 



2 2 

Now, if, as is usually the case, the section where the great- 
est bending-moment acts is also subjected to the greatest 
twisting-moment, it will only be necessary to put for M x the 
greatest bending-moment, and for M 2 the greatest twisting- 
moment. 

§ 298. Thick Hollow Cylinders subjected to a Uniform 
Normal Pressure. — Let inside radius = r, outside radius = 
r iy length of portion under consideration = unity, intensity of 
internal normal pressure = P, of external normal pressure 
=iP v 

i°. Divide the cylinder into a series of concentric rings; 
let radius of any ring be p, and thickness dp, these being the 
dimensions before the pressure is applied. 

Let P become p + £, and dp, dp + dg, after the pressure is 
applied. 

Then at any point of this ring we shall have, for the strain 
in the radial direction, 

f; (0 

dp 

and, since the length of the ring before the application of the 
pressure is 2irp, and after is 2tt(p + $), hence the strain in a 
direction at right angles to the radius is 

p* = * (a) 

2irp p 



894 APPLIED MECHANICS. 

2°. Impose, now, the conditions of equilibrium upon the 
forces exerted by the rest of the cylinder upon the upper half- 
ring. For this purpose let 

/ = intensity of normal pressure on inside ; i.e., at dis- 

tance p from the axis. 
p -\- dp = intensity of normal pressure on outside ; i.e., at dis- 
tance p + dp from the axis. 
Then we shall have for these forces, — 

{a) Upward force due to internal pressure, 

2/(p + £)• 

{b) Downward force due to external pressure, 

2{p + dp)( P + £ + dp + d£). 

(c) Upward force at right angles to radius acting at division 
line between the two half-rings, 

2t(dp + d£), 

where t = intensity of hoop-tension per square unit; i.e., of 
tension in a circumferential direction. Then we have 

2(/ + dp) (p + £ + dp + d£) - 2 p(p + () - 2t(dp + d£) = o; 

and, if this be reduced, and the terms 

2pd£, 2%dp, 2 dp dp, 2dpd£, and 2td% 

be omitted, all of which are very small compared with the 
remaining ones, we shall have 

dp p - t 

/ + - = °- (3) 

dp p 

Now, the two stresses / and t are principal stresses, since 



HOLLOW CYLINDERS SUBJECTED TO PRESSURE. 895 

there are no shearing-stresses on these planes. Hence we have, 
from equations (1) and (2), § 282, 

4* = > - £> < 4) 

E- = t - A (5) 

Now eliminate / and ^ between (3), (4), and (5), and obtain 
a differential equation between p and £ 
Proceed as follows : — 
From (4) and (5), 

= .**_/* + 1_\ 

m 2 — i\tfp mpj 
dp _ E??i 2 ld 2 £ 1 d£ £\ 
dp m 2 — i\dp 2 mp dp mp 2 / 

From (4) and (5) also, 

p — t Em (dB & 



1 Up p/p. 



/0 *» -f- 

Hence, substituting in (3), and reducing, we obtain 

€i + }*i_i =0 (6) 

dp 2 p dp pz N ' 



^jo 2 \p dp p 2 / dp 

Hence, by integration, 

f=-i+ 2 a; ( 7 ) 

dp p 

2a being an arbitrary constant, to be determined from the con- 
ditions of the problem. 



896 APPLIED MECHANICS. 

From (7) we obtain 

* . t Alp) 

Hence, integrating, we have 

£p = a P 2 + J, (8) 

£ being another arbitrary constant. 
From (8) we obtain 

£ = a? + -, (9) 

P 

which gives us, for the two strains, 

y = a -~> ( IO ) 

4> p 

£ • ^ 

- = * + -? (11) 

p p 2 

Hence, substituting these values in (4) and (5), and solving 
for/ and t successively, we obtain 

Em Em b / \ 

p — a , (12) 

m — 1 m + 1 p 2 

Em . Em b , x 

/ = a H . (13) 

m — 1 m + 1 p 2 

Now, to determine # and £, we have the conditions, that, 
when p =z r, p — P, and when p = r iy p =. —P x . 
Hence 

r, Em Em b D Em Em b 

P = a , P L = a — 



m — 1 w + 1 r 2 m — 1 


w + 1 ^i 2 


_ m — 1 /^r, 2 — /V 2 . _ m -\- iP, 


- P 


Em r* — r 2 is#* r x 2 


— r 2 2 ' 


_ ^r, 2 - Pr 2 1 (/> - P)r 2 rf 
r, a — r 2 /o a r, 2 — r 2 


CM; 



HOLLOW CYLLNDERS SUBJECTED TO PRESSURE. 897 

The greatest value of t 9 and hence the greatest intensity of 
the hoop-tension, occurs when p = r; and hence we obtain 

2P 1 r l 2 — P(r* -f r 2 ) g N 

Max / = — — 9 -» (15) 

this value of t being negative when there is hoop-tension, be- 
cause the signs were so chosen as to make t positive when 
denoting compression. 

If P 1 = o, i.e., if there is no external pressure, we have 



u ^'=- p (£^t)> < i6 > 



and, according to Professor Rankine's method, we should deter- 
mine the proper dimensions by keeping max t within the work- 
ing-strength of the material. 

On the other hand, if we decide that we will keep the value 

of e(-\ within the working-strength, we shall find for this, when 

we make p = r, 

\_2Ps 2 - P{r 2 + r 2 )~\ — -P{r* — r 2 ) 
Max^/A = —2 • (17) 

w (^ 2 - r2 ) 

and, if m = 4, 

W ( r r 2 - r *) 

When P, = o, 

Practical cases of thick, hollow cylinders subjected to a uni- 
form normal pressure occur in hydraulic presses and in ord- 
nance. 



898 APPLIED MECHANICS. 



§ 299. Rankine's Theory of Earthwork. — For a mass of 
earth bounded above by a plane surface, either horizontal or 
sloping, his theory assumes the following proportions to be 
true, viz. : 

" i°. The pressure on a plane parallel to the upper plane sur- 
face (which may be called a conjugate plane) is vertical, and 
proportional to the depth. 

" 2°. The pressure on a vertical plane is parallel to the up- 
per plane surface, and conjugate to the vertical pressure. 

" 3 . The state of stress at a given depth is uniform." 

If we let w denote the weight of the earth per unit of 
volume, x the depth of a given conjugate plane below the 
surface, the inclination of the conjugate plane, then is the 
intensity of the vertical pressure on the conjugate plane 

p = wx cos 6. (1) 

Moreover, if is the angle of repose of the earth, then is 
the angle of greatest obliquity. If now we denote by p the 
pressure against the vertical plane, then we have that 



j. ^ n cos 6 — V cos 2 — cos 2 

r >_ wx cos u — , 

cos 6 -\- V cos 2 — cos 2 

j ^ ^ .cos 6 + V cos 2 6 — cos 2 
ana p ^ wx cos tf — - . 

cos — V cos 2 — cos 2 ' 



(3) 



i.e., p may have any value between these two. 

When the problem is to determine the pressure exerted by 
a mass of earth against the vertical face ab of a retaining-wall, 
we determine the intensity of the pres- 
sure/ against the face (acting along cd) 
at a depth x below the upper surface ac 
by means of equation (2) ; inasmuch as 
Rankine claims to prove by means of 
Moseley's principle of least resistance 
that the pressure against the vertical FlG - 3°s. 

plane is the lesser of the two conjugate pressures. 




RANKINGS THEORY OF EARTHWORK. 

Hence, for the entire pressure against the wall we have a dis- 
tributed pressure whose intensity is zero at a, and which varies 
uniformly as we go downwards, the direction of the pressure 
being parallel to the upper surface ae> and the point of appli- 
cation of the resultant being at a depth below a equal to \{ab). 

When the upper surface ae is horizontal, we have cos 6 = i ; 
and hence (2) becomes 

> 1 — sin 

P ^_WX : r— - , (4) 

I -j- sin V ^ 7 

, ' . , 1 + sin . x 

and (3) becomes p — wx : — - . (5) 

SUPPORTING POWER OF EARTH FOUNDATION ACCORDING TO RANKINE. 

Let the surface of the ground be horizontal. 

Then Rankine says that the conjugate pressure may be in- 
creased beyond the least amount by the application of some 
external pressure, as the weight of a building founded upon 
the earth ; that, in this case, the conjugate pressure will be the 
least which is consistent with the vertical pressure due to the 
weight of the building, and if that conjugate pressure does 
not exceed the greatest conjugate pressure consistent with the 
weight of the earth above the same stratum on which the 
building rests, the mass of earth will be stable. 

Moreover, the greatest horizontal pressure at the depth x % 
consistent with stability, is 

1 -f- sin . . 

p = wx — t— 5 • (6) 

1 — sm ' 

The greatest vertical pressure, consistent with this horizontal 

pressure, is 

, 1 -f- sin / 1 + sin 0V ■ x 

/=/ • ^ = wx[— T . \ \ ; (7) 

r 1 — sm \ 1 — sm 0/ w/ 

and this is the greatest intensity of the pressure consistent 
with stability, of a building founded on a horizontal stratum 
of earth at a depth x, the angle of repose being 0. 



900 



APPLIED MECHANICS. 



§ 300. Strength of Flat Plates. — In this regard, the for- 
mulae that will be deduced are those of Professor Grashof, the 
reasoning followed being substantially that given by him in his 
" Festigkeitslehre." 

ROUND PLATES. 

Let the curved line CA be a meridian curve of the middle 
layer of the plate after it is 
bent. Take the origin at O ; 
let axis OZ be vertical, and axis 
OX horizontal, and let the axis 
at right angles to ZOX be (9$, 
so that z, x, and </> are the co- 
ordinates of any point in the 
middle layer of the plate. 

Let y denote the (vertical) 
distance of any horizontal layer 
from the middle layer of the 
plate. 

Let R = radius of curvature 
of meridian line at any point 
(x % z). 

Let R t = radius of curvature of section of middle layer 
normal to meridian line. 

Then we should have, from the differential calculus, 




Fig. 30Q. 



I 

R 



dx 2 



(-+(i) 



d'z 
-^ nearl * 



I 

R, 



1 dz 
x dx 



Hence, reasoning in the same way as in the common theory of 
beams, we should have, for the strains of the layer whose dis- 



STRENGTH OF FLA T FLA TES. 901 

tance from neutral layer is y at point (x, z), provided there is 
no stress in the plane of the neutral layer, 







*x = 


±3? H 


-**■ 






When 


there is 


such 


a stress. 


, let the 


strains 


due to that 


stress be 


t Xo and ^ . 












Then 


we shall have 


= **o ~ 


d 2 z 




(0 






€ * 


= Ho - 


y dz 
x dx 




(») 



Hence, substituting in (1) and (2) of § 283, we have 
mE 



x — 

m 2 



i[^° + <*° - \ m % + 1 £)} (3) 

mE r . fd 2 z . m dz\~] , x 

'♦ = ;f~ r|> + *»*. - y{^ + - j x )\ (4) 

Now let us suppose the plate to be subjected, before load- 
ing, to a uniform pull in its own plane, and normal to its cir- 
cumference ; and let the intensity of this pull be/ x . Then 

and hence, we have, 

€ *° = C *° = ~~mE (5) 

Therefore, substituting in (3) and (4), and reducing, 

mEy I d 2 z 1 dz\ 

°' = P*-^---i\ m l& + - x T x ) < 6 > , 

mEy (d 2 z m dz\ 




902 APPLIED MECHANICS. 

These equations express the stresses in terms of the co- 
ordinates of the points. 

Now impose the conditions of equilibrium upon the forces 
acting on any half-ring of thickness dx = d<f>. 
These forces are — 

i°. Force exerted upon it by the outer part _ .^$*_x 

of the plate, 

\2Xd x -h 2d(X(T x ) \dz. FigTsio. 

2°. Force exerted by the inner part of the plate, 

— 2X0-^. 

3°. Force exerted upon it by the other half-ring, 

— 2<T$dxdz. 

4°. Force exerted by resistance to shear on top and bottom, 

\ (r -f dr) — t 1 2xdx. 

Hence, equating to zero the algebraic sum of these, and 
reducing, we obtain 

dr dr o\£ i d(xo- x ) 



dz dy x x dx 



(8) 



Now substitute for <r x and <r* their values, and reduce, and 
we have 

dr m 2 Ey /d^z , i d 2 z i dz\ , . 

H rr :t- (9) 



dy m 2 — i\dx 2 



i d 2 z i dz \ 
x dx 2 x 2 dx) 



Integrate with regard to y, and we have, since the quantity 
in brackets is not a function of y, 

m 2 Ey 2 (d^z i d 2 z i dz\ 



+ ^ 



2{m 2 — i)\dx 3 x dx 2 x 2 dx) 



STRENGTH OF FLAT PLATES. 903 



h 

But, when^ = - (k being the thickness of the plate), t = o, 

since there is no shearing-force at top or bottom ; 

_ m 2 Eh 2 t d*z 1 d*z _ i_ dz\ 

S(m 2 — i)\dx* x dx 2 x 2 dx) 



T = 



m 2 E(h 2 - 4y 2 ) ( diz 1 d*z _ t_ dz\ ( . 

8(m 2 — 1) \dx* x dx 2 x 2 dx)' ^ ' 



This gives us the intensity of the shearing-stress at any 
point (x, z) at distance y from middle layer ; and this is the 
intensity of the shear at that point between two horizontal 
layers, and hence also along a vertical plane through the point 
(x, z). 

Now let us take the case of a centre load P combined with 
a distributed load / per unit of area. Then shearing-force at 
distance x from centre = 

TTX 2 p + P, 

this tending to shear out a circular piece of radius x. Hence 
we must have this balanced by the whole shearing resistance 
on the surface subjected to shear; 



_ 

.*. 2ttX I rdy — Trx*p + P 

2 
h 



Now substitute the value of t from equation (10), integrate, 
and reduce, and we obtain 



dH 1 d 2 z 
dx 3 x dx 2 



1 A = 6(m* - 1) / P\ 

x 2 dx m'Efc V t*/ 



904 APPLIED MECHANICS. 

Hence, for the intensity of the shearing-force, we have 

■z k 2 - AV 2 / P\ 

4 A 3 \ ttx) 



T = 



This gives the intensity of the shearing-force at any point of 
the plate. 

Next, to find its deflection, or the equation of the meridian 
line, we have, from (12), 

Up. + 1 *) = _ 6 ^ - ^{ px + *) 

dx\dx 2 x dxj m 2 Eh z \ ttx J 

d 2 z 1 dz 6(m 2 — 1) x 2 6(m 2 — 1) P. 

• _l_ == ^ i '—ft __: L. jog x -4- C 

dx 2 x dx m 2 Eh> 2 m 2 Ek 3 -n- 

d 2 z dz 6(7n 2 — 1) x* 6(m 2 — 1) P . 

.'. X = i Lp ^ _ — L-x\og e x + CM. 

dx 2 dx m 2 Efr 2 m 2 Eh* tt 



But 



d 2 z dz 
dx 2 dx 



dx\ dx) 7 



hence, integrating, we have 

dz 6(m 2 — 1) x* 

X 'dx~ m 2 Eh> P ~Z 

6(m 2 - 1) Px 2 , 6{m 2 - 1) P x 2 ; ex* , 

~ m 2 Eh> ^ 1q &* + m 2 Eh* ^7 + V + * («4> 

Hence, dividing through by x, and integrating, 

dim 2 — 1) # 4 
2 == i -p — 

tn 2 Efc 32 

_ «15^1 ^ ( i og ,* - + £ + -flog,* + «.- (15) 

and this is the meridian line of the surface, the constants c % a, 
and e being as yet undetermined. 



STRENGTH OF FLAT PLATES. 905 

This is as far as we can proceed before taking up special 
cases. 

(a) Full Plate. — When the plate is full, the slope becomes 
zero, for x — o; therefore (14) gives us 

d = o, 

and in this case (15) becomes 

_ 6(m 2 - 1) x* 
m 2 Efc 32 

6(m 2 — 1) Px 2 ,. N , ex 2 , , N 

~ azrz,, - "Gog** - + — + e > ( l6 > 

m 2 E/i i it 4 4 

(a) Uniformly Loaded, no Centre Load. — P = O' 

,\ s = - y / 1 -f *. (17) 

m 2 Efo * 32 4 x #/ 

But when ^r = r, ^ = o ; 

6(> 2 — 1) r* cr 2 
""" * = m 2 Efc ^$2 ~~ If 

/6(> 2 - 1) * 2 + ^ 2 \/^ 2 — aA 

•'• * = \^mr-i>^- ~ c )\-T-\ < l8 > 

/2te d 2 £ 

And, substituting for z, — - and -— , their values in (1) and (2), 

ax ax 2 

we obtain 

m — 1 „/3 6(w 2 — 1) A , x 

*• = -IT'* + *(l ^ ^ " ?>• < x »> 

„ tn — 1 /i 6(w 2 — 1) A , v 



and (13) gives 



xh 2 — Ay 2 m „ 



906 APPLIED MECHANICS. 

ifi) Supported all around. — When x — r, <r x = o- Xq = p t for 
all values of y: therefore, from (6), 



\dx i ) r r\dx) r 



dz d* ' z 

and, substituting the values of — and — as determined by 

dx dx 2 

differentiating (18), we have, after reducing, 

c = $ ( m ~ T )(3 m + 1 ) P r * 
"2 m 2 Eh* 

Hence equation of meridian line is 



7. m 2 — 

z = 4 — 

16 m 2 



ii-( 5» + V ^l(f..4 (22) 

EhH m 4- 1 ) 



Hence we have maximum deflection by making x = o ; 

. _ _ 3 (»» ~ i)(5 OT + T ) P r * f 9 ~\ 

" Z °-i6 m 2 Eh> Kn) 

And, substituting in (19) and (20), we obtain, after reduction, 

Et x = — p s + 5 — £ r , ^ 2 - 3^ 2 [^ (24) 
^ = — -a + * — — {-\ 6 ^ r2 - x \y- ( 2 5) 

m 4 w 2 fa [ m + 1 ) 

But, in a plate supported all around, p x = o ; and then the 

h 
maximum value of either one occurs when y = -, and hence 

2 

^.j C- y + o g: {a6) 



STRENGTH OI< *LAT PLATES. 907 

On the other hand, r becomes greatest when x = r and 

y = o. Hence 

3 r 
Maxr = - ip; 

and, if e, represent the maximum strain due to this shearing- 
force, we have 

Max (E€ t ) = [— ^-yCmax r) = - —^- % p. (27) 



RESULTING FORMULA FOR PLATE SUPPORTED ALL ROUND. 

Max Ee Q = * ± ^ — ^— !- —p or £ — — A 

whichever is greatest. 

3 O - i)(5 OT + 1) Z** 4 



2o = 



l6 0Z 2 .£# 



PLATE FIXED AT ENDS. 

Equation (17) applies to this case also. 

Now, when x = r, — — o ; 
dx 

-z n? — 1 , . 

/. c = * pr 2 

2 m z Eh* * 



16 ni* Efc K J v ' 



Hence greatest deflection is 



3 w 2 — 1 /r 4 
*° ~~ 16 m* Eh*' 



908 APPLIED MECHANICS. 



and 

_, m — i , * w 2 — i p , ^ ,x .v 

** = -^-^ + 4 ~^ h^ ~ 3X)y ' (29> 

Eh = «L^i A + 1 f^l £(,* _ ^. (3o) 

m 4 m 2 /i^ 
When p z is positive or zero, then Ee x is maximum for x = o 
j/ = — , and for ^ = r, y = ; and i^ is maximum for 

2 2 

.jr = o, y = - : and the maximum value of iTe^ is equal to first 

2 

maximum of Ee x . We have 

r>- , z? #* — i . . 3 w 2 — i r 2 , 

First max-Ee* = / x + g r - -/, (31) 

^ S m 2 /i 2 

ffl j -? ^ 2 — \ T 2 

Second max^ = /, + ^ — - —p. (32) 

0/ 4 «r h 2 

Hence the second is the real maximum. 



RESULTING FORMULA FOR PLATES FIXED AT THE ENDS. 
Max^ €o = p x 4- — -r 2 P, 

$ m 2 — 1 pr 4 

Z ° " 76 ~^ 2 i£F 

For /, = o, 

r- %m 2 — if*. 

Max^ €o = * — -A 

4 w 2 ^ 2 

§ 301. Thickness of Plates. — Grashof advises the use of 
3 as value of m. If this be adopted, we should have, for the 
proper thickness of round plates, 

Supported. Fixed. 



V " - ' v 3/ 



RECTANGULAR PLATES. 



9O9 



where h = thickness, r .= radius,/ = pressure per square inch, 
and f = working-strength per square inch. If, now, we use a 
factor of safety 8, and use as tensile strength of cast-iron 20000, 
of wrought-iron, 48000, and of steel 80000, we should have : — 





Supported. 


Fixed. 


Cast-iron . . . 
Wrought-iron 
Steel .... 


h = o.oi&2$ h }orSfp 
h = 0.0117850?-^ 
h = o.oogi28'jr\p 


h = 0.016 3300^^ 

h = 0.0105410/^/ 

* h == 0.0081649?-^ 



§ 302. Rectangular Plates. — Refer the plate to rectangu- 
lar axes, as before, OZ, OX, O® ; the origin being at the middle- 
of its middle layer. 

Let y = distance of any point in the plate from the middle 
layer. 

Let p x be the radius of curvature of a normal section par- 
allel to OX at the point (x, z, <f>). 

Let pt be the radius of curvature of a normal section par- 
allel to O® at the point (x, z, <f>). 

Then we shall have, by the principles of the common theory 
of beams, 



€ x = «*- ± 



H — H ± —> 
p$ 



where e Xo and c^ are the strains of the middle layer in the 
directions OX and O® respectively. 

Moreover, from the Differential Calculus, we have 



Px = 



v- + (IT + (IT 



d 2 z 
dx* 



cos 2 A 



d 2 z x • d 2 z , ' 

: cos A cos ik H cos 2 u 

dxd<j> r d<f> 2 



9IO APPLIED MECHANICS. 



where A = angle between normal and z axis, and ^ = angle 

dz dz 

between normal and x axis. But — - and — being the slopes, 

dx a<f} 

and hence small, we shall have nearly 

cos A = i, cos /x = o, 



i d 2 z i d 2 z 
p x dx 2 p<f, d<$? 




d 2 z 

*x = * x — V » 

° dx 2 


(I) 


d 2 z 

H = u> n — y-r—- 
° d(f> 2 


(2) 



Hence (i) and (2) of § 283 give us 

mE , N mE ( d 2 z d 2 z ) 

<r* = — — ( m€ * + «*<,) - y— — \ m ^- 2 + -7^ \> :. 

m 2 — 1 a» 2 — 1 ( dx 2 d<f> 2 ) 

mE . N mE [ d 2 z , d 2 z ) 

o-a = U x + *»«* ) — y { - h m— — > . 

* w 2 - 1 ° *° w 2 - i(<£c 2 aty 2 ) 

And, if o- Xq , <r^ o , denote the stresses in the middle layer, we 
shall have, since 

mE , x mE . , x 

m 2 E \d 2 z , 1 d 2 z) 
-m^^ y \dx 2 + md+ 2 \> <3> 

( 1 d 2 z , ^ 2 s ) , . 

1 ( /# ^r 2 tf <£ 2 ) 



m 2 E [ 1 d 2 z . ^ 2 z 



Now, if £ and ^ denote the increments in x and <f> respec- 
tively due to the load, we shall have 

C* j dz dt, d 2 z 

£ = / €xdx = x€ Xo - y— .*. — = -y- 



-£ 



dx d<f> dxd<f> 

,, , dz din d 2 z 

H d<j> = $ Ha - y— .'. -i = ->>- 



<fy dfr akaty 



RECTANGULAR PLATES. 9II 



But 



hence 



T ** =G ™= G (% + i)' 



T * = ~ 26 W (5) 

Equations (3), (4), and (5) are the expressions giving the 
stresses on two planes at right angles to each other, parallel to 
OX and O® respectively. Hence we have a case of stress on 
two planes at right angles to each other, and we are to find the 
principal stresses : we thus have — 

i°. Normal stress on x plane, <r x . 

2°. Shearing-stress on x plane, t^. 

3 . Normal stress on <f> plane, cr . 

4 . Shearing-stress on <f> plane, r^ Xm 

Hence, if we denote by <r z and <r 2 the maximum and mini- 
mum principal stress, we have (§ 290) 



0-1 = i(o* -f- 0-4,) -f #I{<t x - °-<t>) 2 + 4^ 2 , (6) 



o- 2 = £(°* + <ty) - i\f(<r x — v+y + 4tVJ (7) 

and hence, if e, and « 2 denote the strains in the directions of 
the principal stresses, 



<r 2 m — 1 



m -\- 1 



+ — ^T-V^K - **) a + 4^, (8) 



e, o-j m — 1 / . 

A^ = cr 2 — — = — —-(a* + 0^) 



W + I 



, VV* - °>) 2 + 4rV; (9) 



912 APPLIED MECHANICS. 

and for the strain c 3 , parallel to OZ, we have 

In order to use (8), (9), and (10), however, we must know 
°"jn °>> an d t^ ; and for this purpose we must know the equa- 
tion of the middle layer after bending. For this purpose, apply 
the equations (1), (2), (3), of § 281 to any particle dxd^dz in 
the interior of the body. We have then, X = Y = Z = o. 
Therefore 



da- x dr xz dr x $ 

dx + dy + d<f> ~ ° 


dr xz 


dv<t, dr x $ dr^ z _ 




dy dcj> dx 





d<j> dx )' 



Therefore, making use of (3), (4), and (5) with the above 
conditions, we deduce 



d(r x m 

dx m 2 



Ey ldhs_ i_ d^z \ 
~^~i\dx^ mdxd^r 



d<T4> — m 2 Ey/d^z 1 d*z \ ( . 

^Atip m dx 2 d<f>)' r I2) 

(13) 



d(f> m 2 — iWc/> 3 
dr X 4> _ mEy d*z 



dx m -f 1 dx 2 d<f> 

dr x ^ _ _ mEy dH . , 

dcf> m -f- 1 dxd<f> 2 

dr xz = m 2 Ey fd^z d*z \ ' . 

dj> #z 2 — 1V&: 3 dxdcfi 2 /' 

dr^z = ;;? 2 ^> /^ 3 , J!±_\ / l6 ) 

4; w 2 - 1 \d<p dx 2 d<f>J 



RECTANGULAR PLATES. 913 

Hence, by integrating (15) and (16), we have 
m*Ev 2 fd*z d*z \ 

TxZ ~ 2(m* - i)W ^^/ '* 
m 2 Ev 2 f d 3 z d*z \ 

T * " 2(^ 2 - l)W ^^/ '"* 

But when v = -, r^ = r xz = o ; 



and 



and 



Hence 



S(m 2 — i)\d& dxd<j> 2 ) 

m 2 Eh 2 (d*z dH \ 

8(w 2 - \)\d^> dx 2 d<f>) 

m 2 E I d*z d*z\!v 2 __ k 2 \ 

•"■ T * 2 - w2 - i\^^ + op) v7 8"; 

= m*E l d*z d*z \lv 2 __ H\ 
Txz ~ W 2 _ j^-^ dxd<f> 2 )\2 8/' 



^ = jr?E_l_d^_ , d*z\tv 2 _ #\ 
d£ "" w 2 - iV*W</> 2 ^/\2 8"/' 



» 2 A 2 8"/ 



d&c ^z 2 — i\d&* dx*d<p 

Now we have <r 2 = /, where / is the intensity of the load ; 
therefore the third equation gives us, on integrating between 

the limits - and — -, 

2 2 

2 2 



gi4 



APPLIED MECHANICS. 



m 2 E f/z>3 h 2 v\(d*z 



^ w 2 - i(V6 



4- 2-^— 4- — V 1 - 

Vw<£ 2 <ty 4 /_*J = 



m 2 E l d*z d*z d*z\l ' fc_ __ kP 

d 4 z , d 4 z , d*z J2(m 2 — i)/> 

-r~. + 2— — T— + 



^3 _ £\ 

2 4 8y~ 



dfc* tfW<^ ^ 4 



?n 2 Efa 



(»7> 



and this is the differential equation of the surface, and should 
be integrated in each special case. 



INDEFINITE PLATES WHICH ARE FIRMLY HELD AT A SYSTEM OF 
POINTS DIVIDING THEM INTO RECTANGULAR PANELS. 

Let the sides of the panels be 2a and 2b. Assume the 
origin at the middle of the panel, the axis of x being parallel 
to 2<z, and the axis of y parallel to 2b. We shall in this case 
have the following conditions ; viz., — 

dz 

(a) — = o for x = zka and all values of <L 
dx 

dz 

(b) — = o for <f> = ±b and all values of x. 
d(f> 

(c) z = o when x = zbtf, <f> =■ ±:b. 

{d) If we develop the value of z in powers of x and <f>, there 
must enter only even powers of x and <f>, since the value of z 
remains the same when we put —x for x, or — <f> for <£. 

Now, if we write 

z = A + Bx 2 + C<f> 2 + Dx 2 <f> 2 + Ex* + F<p 

+ Gx b + Hx^ 2 + Kx 2 <f>* + ^* 6 + M&> etc -f 



the above conditions will be fulfilled : — 



RECTANGULAR PLATES. 915 

i°. By making all the co-efficients after the fourth, each zero. 
2°. By making D — o, therefore writing 

z = A + Bx 2 -f- C0 2 + Ex* + F<j>\ 



Now 



— = 2Bx + 4^^3, ^- = 2^ + 4F<f>\ 

dx dd> 



and 

o = ^ + ^^ 2 4- Cb 2 + ^* 4 + ^ 4 

.-. B = -2Ea 2 , C = -2Fb 2 , 

.-. A = 2Ea* + 2i^4 _ ^4 _ pfr = Ea* -f /#*. 

Hence the equation becomes 

z = Ea* + 7^4 _ 2 ^^^ 2 - 2i# 2 <£ 2 + Ex* + j^4 

= ^(« 2 - x 2 ) 2 + ./?(£» - <f> 2 )* 

dz 
•\ — - = — ^Exia 2 — x 2 ) = 4Ex* — AfEcPx 
dx 



also 



- — = \2Ex 2 — 4 J £'« 2 , 
dx 2 



dz 

— = -4E<j>(b 2 - <f) = 4^3 - 4 Fb 2 <j> 

d<j> 



... g= I2 ^- 4 ^, 

*/ 3 z */ 4 Z */ 3 £ ^ 



dx 2 dc}> dx 2 dcf> 2 dxd<j> 2 dx'd^ 2 

d*z jr. d*z _. ^* ^ 

_ = 24^ — = 24^, — = 24^, — = 24^ 

.-. 2 4 (^ + jo = I2( ^ 2 ~ T)/ .•: E + F= iz*-j)£. 

Hence equation of the middle layer is 
z = E(a* - x*y + F{b 2 - <£ 2 ) 2 , where E + ^ = ^ ~ *^ . (18) 



giQ APPLIED MECHANICS. 

Now, in the case of an ordinary beam fixed at both ends, 
and loaded uniformly with p lbs. per unit of area, if b is the 
breadth, we have : — 

i°. The points of inflection are at a distance from the middle 

equal to —=, where a is the half-span ; and 
V3 
2°. The bending-moment at a section at a distance x from 

the middle is — ( — — x 2 \ when x <-=, and^—lx 2 ) when 

2\3 / V3 2\ 3/ 

X > — ; therefore the value of z is found from the formula 



*=AOV-3-'*"*>^ 



3 

or 

• = MTHV + &nTi - -•)-•*» * < % 

Either one, when integrated, gives for z the value 

2Ete 

Hence in the flat plate, if b = o, the values of E and F must 

be such that the formula shall reduce to z = tl. (a 2 — ^tr 2 ) 2 

when b — o. Now, it does reduce to z — E{a 2 — oc 2 ) 2 . There- 
fore E must be such a function of a and b, that, when £ — o, it 

shall reduce to -?—. So likewise F must be such a function 

2£V^ 

of a and £, that, when a = o, it shall reduce to — ^— . Suppose, 
then, we put 

E = -£- + *v and ^ = -^rr + *% 

2 Eh* 2 Eh 3 

since these functions fulfil the above conditions. 



RECTANGULAR PLATES. 917 



Now we have 



E + F = (™ 2 ~*)r 
2m 2 Eh* 



Eh* V ' 2W^A3 



A («" + *■)' = - * ^r~ 

2m 2 Eh* 

(m 2 + i)p 






2m 2 Eh*{a n + £") 



dfc*' 



_. 1 jji*z 

He* = <r<* ov — yxi— — • 

* *° m ° ' d<f> 2 

d 2 z d 2 z 

Hence, substituting for — — and -7— their values, and observ* 

## 2 #0 2 



ing that 



c* is greatest for x — ±a, y = ±~, 

2 



€0 is greatest for <f> = ±£, jy = ±_, 

2 



we obtain 

I 991* CL* 

max (Ee x ) = ov. o> o ± 2 — —a (19) 



»2 ^2 



max <*„) = <r 0o - ^ ± a an + m l |/. (*>) 



Ql8 APPLIED MECHANICS. 



These may be written as follows 



max Et x = (t Xq — — o> o ± 2 



1 - -f £ V 

m\a) a* . v 



m i 4. (iY h2 
l/^V 

77 I , m\b) b 2 , , . 

max^ = <r fe - -^ ± ■ ^ + Xi jj. (22) 

We have also, by substituting for £" and F their values in 
equation (18), 



f a* ^b» b n —a n 

, = _iL_ %-(& - x*)* + ^-(^ 2 - <£ 2 ) 2 I . ( 

2^/^ 3 I a" + b n v y a* -J- 3* v ^ J i v 



23) 



In these results the exponent n is undetermined, and we 
have no means of determining it in the general case. We only 
know, that, since the deflection must increase for a decrease in 
x and <f>, therefore we must have, whenever a > b, 

a\ M ^ ^ 2 ^ 2 log m 



, < m 2 ,\ n < 



K!) 



This leaves the general case indeterminate ; but a common 
practical case is not subject to this indetermination, i.e., the 
case when a = b, for then 



©-=©■- 



whatever the value of n; and hence equations (21), (22), and 
(23) give 

max v^) = <r x o> ± — —p, (24) 

w #z 2 /? 2 

T /?Z 2 - — T (J? 1 

max &, = <r* o - -o^ ± —j— -/, (25) 



and 



RECTANGULAR PLATES. 919 



Z m ^=-1 -£-\ (a* - X 2 ) 2 + (« 2 - &Y\, (26) 



max z = — p. (27) 

m 2 lEh* 



FORMULA FOR THE . SHEETS OF A LOCOMOTIVE FIRE-BOX. 

In this case we have a = b ; hence (24), (25), and (27) 
apply : and if we write, with Grashof, m = 3, they become 



1 8 a 2 

max (Et x ) = or^ - - <r* o + - -/, (28) 

max (.£«*) = <r +o - i ov c + - ^», (39) 

max (z) = J m (30) 

Now, in the case of the horizontal sheets, <t Xq = <r^ o = o, 
and we have 

max (Ee x ) = - j 2 p, (31) 

maX ( " } = 9 U* (32) 

In the case of the vertical walls, inasmuch as these have to 
resist the steam-pressure in a vertical direction, the inner one 
is called upon to bear compression, and the outer tension, in a 
vertical direction. If / is the length of the outside of the fire- 
box, and /, its breadth, we shall have for the outer plate, taking 
axis of x vertical, 



920 APPLIED MECHANICS. 

and for the inner plate, if / and // are corresponding dimen- 
sions of inside of fire-box, 

And, by making these substitutions in (28), (29), and (30), we 
obtain our formulae. 



RECTANGULAR PLATE FIXED AT THE EDGES. 

For this case Grashof deduces the equation of the middle 
layer as follows : 

i°. This equation must be a function of x and <j>. 

2 . If 2a and 2b are the sides of the plate, this function 
must become 

(a) When b = 00 for all values of <f>, 

(/?) When a = 00 for all values of x, 

— ■&&•- rr> , 

because the plate then becomes a beam fixed at the ends. 
The function that will satisfy these two conditions is 

^ = p (ga _. ^y^ _ py 

2Eh> a* -f M ' K ' 

From this he deduces for max 2, when x = $ = o, 

1 p aW g . 

2 Eh* a* + M x ' 



RECTANGULAR PLATES. 92 1 

From (1) he deduces 

d 2 z 2p (a 2 — $x 2 ) (b 2 — <t> 2 )* 

(4) 
(5) 



dx 2 


Efc a* + b* 


d 2 z 


2p (a 2 — x 2 ) 2 (b 2 — $<f> 2 ) 


d<f> 2 


Eh* a* + M 


d 2 z 


8p (a 2 — x 2 )x(b 2 — <f> 2 )<f> 



dxd<f> Eh* a* + b* 

d 2 z Ap a 2 b* c « . ,,- 

max-— = %- . , , . , for * = ±«, tf> = o, (6; 

d 2 z Ap a*b 2 e . , , , * 

max—- = -2£ -, for <£ = ±0, a; = o, (7) 



^ 2 2 
max 



= 32 _/_ _^3_^ for ^ = 1^ ^ = l^ (g) 



dfoaty 27 ^ #4 + 34 3 3 

these corresponding to the points of inflection of a loaded 
beam fixed at the ends. 

Hence (1), (2), and (5) of § 300 give 

Z7 I 2 ^ 4 a * J. / \ 

max -fie* =o-^ cr^ ± *>, (o) 

max (i?e<*>) = a* ov ± -A (10) 

, x 8 m 2a 2 b 2 ab M , v 

max (r 2 ) = — /. (11) 

At the places where e* and c^ are greatest, 

T* = o. 
At the place where r z is greatest, 

Hence it is either (9) or (10) that gives us the suitable 
formula to use in any special case. 



922 APPLIED MECHANICS. 



EXAMPLES OF THEORY OP ELASTICITY. 

i. It has been sometimes proposed to use oblique seams in a boiler- 
shell. Assume the seams at an angle of 45 ° with the axis of the boiler, 
a pressure of 100 lbs. per square inch of the steam, and a diameter of 
4 feet. Find the tension per inch of length of seam, and its direction. 

2. Given a shaft carrying 80 HP, and running at 250 revolutions 
per minute. Suppose the driving-pulley to be at the middle of the 
length, this being 6 feet, and given that the ratio of the tension on the 
tight side of the belt to that on the loose side is 3.75. Find the proper 
size of shaft, assuming 10000 lbs. per square inch as the working-strength 
of the iron. 

3. What should be the thickness of a flat plate to bear 150 lbs. 
pressure per square inch, and stayed at points forming squares 8 inches 
on a side, the plate being of wrought-iron, working-strength 10000 lbs. 
per square inch. 

4. Find inner radius of a hydraulic press to bear 1500 lbs. per 
square inch, given outer radius =18 inches; material, cast-iron; ten- 
sile strength 20000 lbs. per square inch. 






INDEX.. 



PAGE 

Acceleration ^. 75 

Angular momentum 106 

Arches 779 

conditions of stability 800 

correcting joints 811 

criterion of safety 801 

criterion of stability 818 

elastic 827 

general remarks 825 

linear 789 

line of resistance 795 

line of resistance determined by two 

points. . 804 

modes of giving way 790 

Scheffler's method 805 

true line of resistance .• 803 

unsymmetrical arrangement 819 

Atwood's machine 79 

Axes of symmetry of plane figures 115 

Bar-iron, tests by Kirkaldy 394 

Bauschinger, building-stones 717 

cast-iron columns 372 

cement 733 

repeated stresses 531 

timber 7°7 

Beams assumptions of common theory. 268 

cross-section of equal strength 298 

deflection of 299 

deflection with uniform bending mo- 
ment 306 

fixed at the ends 312 

load not at middle 307 

longitudinal shearing 319 

mode of ascertaining dimensions. ... 292 

mode of ascertaining stresses. 291 

modulus of rupture 293 

moments of inertia of sections 275 



PAGE 

Beams, oak 684 

position of neutral axis 269 

principles of common theory 267 

rectangular, slope and deflection. ... 311 

resilience of 306 

shearing-force and bending-moment . 272 

slope and deflection 301 

slope and deflection, special cases. . 302 
slope and deflection under working- 
load 310 

spruce 675 

table of deflections and slopes 305 

table of shearing-forces and bending- 

moments 274 

timber, strength and deflection 673 

timber time tests 688 

uniform strength 296 

variation of bending-moment with 

shearing-force 317 

white-pine 686 

working-strength 293 

wrought-iron ? strength and deflec- 
tion 440 

yellow-pine, strength and elasticity 681 

Beardslee, effect of rest 402 

reduction in rolls, wrought-iron 401 

shape of specimen 400 

tensile limit 400 

tests of wrought-iron. 399 

Bending and torsion combined 889 

and twisting 338 

Bending-moment 185 

in beams 272 

Bending-moments, graphical representa- 
tion 287 

Bollman's truss 219 

Bow's notation 143 

Breaking-strength 245 

923 



924 



INDEX. 



PAGE 

Bridge columns, Clark, Reeves & Co. ... 415 

Watertown -arsenal tests 421 

Sondeiicker's formulae 419 

Bridge-trusses 184 

actual shearing-force 203 

compound 208 

concentration of loads at joints 203 

counterbraces 200 

diagonals 200 

examples 186 

general formulas 209 

general remarks 219 

method of sections 180 

shearing-force and bending-moment 185 
steps in determining stresses under 

fixed load 186 

vertical posts 202 

with vertical and diagonal bracing.. . 198 

Building-stones 714 

Buttress, stability 793 

Cast-iron 356 

columns 364 

composition and characteristics 356 

list of experimenters 358 

tension 359 

transverse strength 375 

Catenary 784 

transformed 787 

Cement mortar 720 

Centre of gravity 221 

Centre of gravity, examples 286 

f a line 225 

of a slender line 224 

of flat plate 223 

of homogeneous bodies 223 

of plane areas 224 

of solid bodies 226 

of symmetrical bodies 237 

of system 221 

Pappus's theorems 234 

Centre of percussion , 1 20 

Centre of stress 263 

Centres of percussion and of oscillation, 

interchangeability 123 

Chain cable 403 

Chain or cord, loaded 779 

Coefficients of expansion 507 

Cold rolling 495 

Collision 103 



PAGE 

Columns, cast-iron 364 

Euler's rules 326 

Gordon's formulae 324 

Hodgkinson's rules 328 

oak -. 653 

strength of 323 

timber 652 

white-pine 659 

wrought-iron 414 

yellow-pine 653, 664 

spruce 670 

timber with bolster 670 

Components of velocities of, and forces 

acting on, a body 82 

Compound bridge-trusses 208 

Compression, direct 253 

of timber 652 

of wrought-iron 413 

Wohler's experiments 254 

Continuous girders 743 

distributed and concentrated loads. . 770 

examples for practice 778 

loads concentrated 763 

loads distributed 744 

Contraction of area, Kirkaldy 394 

Cord or chain, loaded 779 

Cord with load uniformly distributed. 

horizontally 782 

Counterbraces ' . 200 

Couple, composition of, in inclined 

planes 59 

Couples, composition of 57 

effect on rigid body 55 

effect when forces are inclined to 

rod 54 

measure of rotary effect 53 

moment of ' 53 

effect on rigid rod 51 

representation by a line 59 

resultant with single force 61 

Crystalline fracture, Kirkaldy 498 

Crystallization of iron and steel 498 

Cylinders, thick hollow, strength of . . . 893 
thin hollow 251 

Deflection of beams 299, 301 

Domes 843 

Dynamics 75 



Eccentric load on columns. 



331, 366,420 



INDEX. 



92 



PAGE 

Elasticity, modulus for timber beams. . 696 

modulus, cast-iron 360, 363 

modulus for use with timber beams . . 695 

modulus of 240 

modulus, Rosset 363 

modulus, wrought -iron 405, 407, 409 

theory of 852 

Ely, rules for strength of timber posts. . 671 

Energy 78 

Equilibrium curves 7 79 

Euler's rules for commns 326 

Expansion, coefficients 507 

Ewing cast-iron columns 374 

Eye-bars, steel, Watertown Arsenal 487 

wrought -iron 412 

Factor of safety for iron and steel 519 

timber 672 

Pink's truss 217 

Floors, timber 699 

Force 3 

and momentum, relation between .... n 
applied at centre of gravity of rigid 

rod 51 

applied to rigid rod, not at centre. . . 46 

centrifugal 81 

centrifugal, of solid body 85 

characteristics of 16 

criticism of definition 6 

definition of 8 

deviating 82 

external 9 

intensity of, distributed 40 

measure of 5.9,76 

moment of 3° 

relativity of 9 

resultant of, distributed 40 

single, at centre of rigid rod 43 

Forces, centre of system of parallel 38 

composition of 2 ^,2?> 

composition of parallel 37, 62 

composition of parallel, in a plane. . . 36 

composition of two 30 

co-ordinates of centre of parallel. ... 39 

decomposition of 19 

distributed 39 

effect of pair on rigid rod 50 

equilibrium of 28 

equilibrium of, in a plane 69 

equilibrium of, in space 73 



PAGE 

Forces, equilibrium of parallel 37 

equilibrium of three parallel 31 

moment, causing rotation 49 

normal and tangential components . . 80 

parallelogram of 16 

polygon of 23 

resultant of any number of parallel. . 35 

resultant of, in a plane 66 

resultant of, in space 70 

equilibrium of, in space 73 

resultant of two parallel 33 

equilibrium of three parallel 31 

statical measure of 12 

Frames, stability 140 

Frame, triangular 141 

isosceles triangular. 143 

polygonal 145 

Frames of two bars 138 

Frames, stability .- 140 

Framing-ioints 698 

Friction of blocks 791 

Funicular polygon 147 

G, relation to £ 869 

G, value of 870 

Girder, greatest stresses 193 

Girders, continuous .1 743 

distributed and concentrated loads. . 770 

examples for practice 778 

loads concentrated 763 

Gordon's formula? 324 

Grade, effect on tractive force 100 

Gravity, centre of 42, 221 

Half-lattice girder: travelling-load, 

greatest 192 

Hammer-beam truss 176 

wind pressure . 179 

Harmonic motion 102 

Headers of timber 698 

Hodgkinson's rules for columns 328 

tests of cast-iron columns 364 

Hooks, strength of 322 

Impact 123 

central 1 24 

coefficient of restitution 125 

coefficient of restitution by experi- 
ment , 132 

elastic 127 



926 



INDEX. 



TAGE 

Impact, imperfectly elastic 129 

inelastic 126 

oblique 134. 

of revolving bodies 136 

special cases of elastic 128 

Impact, special cases of imperfect elas- 
ticity 132 

special cases of inelastic 128 

velocity at greatest compression. ... 125 



KlRKALDY. 

tests of bar iron. 



394 



Launhardt's formula 248, 524 

Load, sudden application of 246 

Longitudinal shearing of beams 31 q 

Mass, measure of 10 

unit of 13 

Materials, strength of 240 

Metals and alloys other than iron and 

steel 642 

Modulus of elasticity 240 

approximate values 245 

Modulus of rupture, for beams 293 

Moeller, cast-iron columns. . 373 

Moment of deviation 108 

Moment of inertia 106 

Moments of inertia about different axes . 113 

components of 117 

Moments of inertia, equal values of 116 

examples 118 

of Phoenix columns 286 

of plane figures about parallel axes. . no 

of plane surface 107 

of sections 275 

of solids around parallel axes 117 

polar, of plane figures in 

principal - 114 

Momentum n 

Mortar (cement) 720 

Motion and rest 2 

Motion, Newton's first law of 4, 9 

Newton's second law of 13 

01. curved lin« 95 

on inclined plane „ 92 

relativity of 1 

under influence of gravity 87 

•uniform. 76 

•uniformly v:.rving 76 



PAGE 

Motion, uniformly varying rectilinear. . . 87 

Motions, parallelogram of 15 

polygon of j 5 

m, value of 870 

Nails in one pound i 5I 

Neutral axis of beams 269 

Notation, Bow's 143 

Oak columns 654 

Pappus's theorems 234 

Pendulum, cycloidal 99 

simple circular 9 7 

Pig-iron 356 

Plates, flat, strength of 900 

thickness of 908 

Polygonal frame i 45 

Projectile, unresisted 89 

Punching and drilling plates 548 

Radius of gyration 107 

Rankine strength of timber 647 

Reduction in rolls, Beardslee, wrought - 

iron 401 

Resilience of a beam 305 

of tie-bar 247 

Resistance, line of -. 795; 

Resistance, line of maximum and mini- 
mum 799 

to direct compression 253 

to shearing 256 

true line of 803 

Rest effect. Beardslee 401 

Rigid bodies, rectilinear transference of 

force 29 

rotation of 105 

statics of 29 

Riveted ioint.-> 546 

rules by P. Schwamb 550 

transverse strength 555 

punching and drilling 548 

Watertown-arsenal tests 570 

Rodman, strength of timber 650 

Roofs, estimation of load 163. 

weight of materials 151 

Roof-trusses. 138, t 60 

determination of stresses 150, 165 

distribution of load 1 63 

examples 1 63 



INDEX. 



927 



PAGE 

Roof -trasses, general remarks 172 

with loads at lower joints 171 

Rope, wire 639 

Rosset's tests of cast-iron 363 

Rupture, modulus for beams 293 

Scheffler's method with arches 805 

mode of correcting joints 811 

Scissor-beam truss 179 

without horizontal tie 180 

Seasoned columns 655 

Seasoning, effect on timber 712 

Semi-girder, greatest stresses 193 

■ Shafting, strength of 333, 539 

Shafts, transverse deflection 337 

under combined torsion and bend- 
ing 889 

Shape of specimen 351 

Kirkaldy 394 

Shearing-force 185 

actual for bridge-trusses 203 

in beams 272 

Shearing, resistance to 256 

Shearing-strength of iron and steel. 419, 539 

Shearing of timber 711 

of timber beams 695 

Slating, weight 151 

Slope of beams 301 

Slotted cross-head 102 

Snow, weight of 151 

Springs 338 

torsional strength 339 

transverse strength 343 

Spruce beams 675 

Stability of a buttress 793 

of an arch .800 

of position *~ 793 

Standard specifications for cast-iron. . . . 385 

for cement 7 24, 7 29 

for steel 450 

for wrought-iron 395 

Steel, composition, kinds, and charac- 
teristics 445 

crystallization 498 

effects of temperature 506 

eye-bars 487 

tensile strength and elasticity 

459, 464, 487 

torsional strength 487 

wire 487 



page 
714 
240 
857 
852 
855 
865 



Stone 

Strain 

resultant 

Strains 

in terms of distortions 

relation to stresses 

Strength, breaking and working 245 

of columns 323 

of hooks 322 

of materials 240 

of materials, general remarks 350 

of shafting 333 

Stress 240 

centre of 263 

compound 861 

converse of ellipse 884 

ellipse 881 

graphical representation 261 

intensity of 260 

relation to strains 865 



simple 860 

tangential 861 

uniform 264 

uniformly varying 265 

Stress, uniformly varying, amounting 

to a statical couple 266 

Stresses 859 

Stresses, composition of 872 

conjugate 872 

equilibrium of 865 

greatest, in girder 193 

in roof-trasses, determination of 150 

mode of ascertaining, in a beam 291 

parallel to a plane, composition.'. . . . 875 

principal 877 

principal, determination of 878 

Stretching and tearing 242 

Strut 138 

Struts, short 323 

Suspension-rod of uniform strength 250 

Temperature, effect on iron and steel. . 506 

Tensile limit. Beardslee 400 

Tension of timber 651 

Theory of elasticity 852 

Timber beams, immediate modulus of 

elasticity 695 

beams modulus of elasticity for 

use 695 

beams, time tests 6SS 



928 



INDEX. 



PAGE 

Timber columns 652 

compression 652 

effect of seasoning 712 

factor of safety " 672 

floors 699 

framing- joints 698 

general remarks 712 

longitudinal shearing in beams... 695 

posts, rules for strength, Ely 671 

strength as given by Rankine 647 

strength as given by Rodman 650 

tension 651 

tests by Bauschinger 707 

transverse strength 673 

Time of descent down a curve 96 

Time tests on timber beams 688 

Torsional strength of iron and steel 487 

Torsional strength of springs 339 

Torsion and bending combined 889 

Translation and rotation combined 44 

Transverse deflection of shafts 337 

Transverse strength of cast-iron 375 

springs 343 

timber 673 

wrought-iron 440 

Travelling-load 192 

Triangular frame 141 

Triangular truss, wind pressure 147 

Truss, Bollman's 219 

Fink's 217 

hammer-beam 176 

scissor-beam 179 

triangular, wind pressure 147 

Trusses, bridge 184 

methods for determining stresses 140 

roof 138, 166 



PAGE 

Trusses, roof, determination of stresses. . 150 
Twisting and bending combined 338 

Velocity 2, 75 

Watertown Arsenal, steel eye-bars. . . 487 

tests of bridge columns. 421 

tests of riveted joints 546 

Weight of materials for roofs 151 

of snow ^ 151 

Weyrauch's formula 255, 525 

White-pine beams 686 

columns 659 

Wind pressure 152 

triangular truss 147 

Wire, and wire rope 639 

W°h]er's experiments 526 

Working-strength 245 

of beams 293 

Work, mechanical 77 

under oblique force 104 

unit of 7 7 

Wrought-iron beams, strength and de- 
flection 440 

Wrought-ircn characteristics 391 

compressive strength 413 

crystallization 4QS 

effect of temperature 506 

eye-bars 412 

tests by Beardslee 399 

tests by Kirkaldy 394 

transverse strength <-4° 

Yellow-pine beams, strength and elas- 
ticity - 681 

columns, tables 653, 664 



SHORT-TITLE CATALOGUE 

OF THE 

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Mandel's Handbook for Bio-chemical Laboratory nmo, 

* Martin's Laboratory Guide to Qualitative Analysis with the Blowpipe . s nmo, 
Mason's Water-supply. (Considered Principally from a Sanitary Standpoint.) 

3d Edition, Rewritten 8vo, 

Examination of Water. (Chemical and Bacteriological.) nmo, 

Matthew's The Textile Fibres 8vo, 

Meyer's Determination of Radicles in Carbon Compounds. (Tingle.), .nmo, 

Miller's Manual of Assaying nmo, 

Mixter's Elementary Text-book of Chemistry. nmo, 

Morgan's Outline of Theory of Solution and its Results nmo, 

Elements of Physical Chemistry nmo, 

Morse's Calculations used in Cane-sugar Factories i6mo, morocco, 

Mulliken's General Method for the Identification of Pure Organic Compounds. 

Vol. I Large 8vo, 

O'Brine's Laboratory Guide in Chemical Analysis - 8vo, 

O'Driscoll's Notes on the Treatment of Gold Ores 8vo, 

Ostwald'? Conversations on Chemistry. Part One- (Ramsey.) nmo, 

Ostwald's Conversations on Chemistry. Part Two. (Turnbull ). (In Press.) 

* Penfield's Notes on Determinative Mineralogy and Record of Mineral Tests. 

8vo, paper, 

Pictet's The Alkaloids and their Chemical Constitution. (Biddle.) 8vo, 

Pinner's Introduction to Organic Chemistry. (Austen.) nmo, 

Poole's Calorific Power of Fuels 8vo, 

Prescott and Winslow's Elements of Water Bacteriology, with Special Refer- 
ence to Sanitary Water Analysis nmo, 

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* Reisig's Guide to Piece-dyeing 8vo, 25 00 

Richards and Woodman's Air, Water, and Food from a Sanitary Standpoint 8vo, 2 00 

Richards's Cost of Living as Modified by Sanitary Science nmo, 1 00 

Cost of Food, a Study in Dietaries i2mo, 1 oo- 

* Richards and Williams's The Dietary Computer 8vo, 1 50 

Ricketts and Russell's Skeleton Notes upon Inorganic Chemistry. (Part I. 

Non-metallic Elements.) 8vo, morocco, 75 

Ricketts and Miller's Notes on Assaying 8vo, 3 00 

Rideal's Sewage and the Bacterial Purification of Sewage 8vo, 3 50 

Disinfection and the Preservation of Food 8vo, 4 00 

Rigg's Elementary Manual for the Chemical Laboratory 8vo, 1 25 

Rostoski*s Serum Diagnosis. (Bolduan.) nmo, 1 00 

Ruddiman's Incompatibilities in Prescriptions 8vo, 2 00 

Sabin's Industrial and Artistic Technology of Paints and Varnish. ...'... .8vo, 3 00 

Salkowski's Physiological and Pathological Chemistry. (Orndorff.) 8vo, 2 50 

Schimpf's Text-book of Volumetric Analysis nmo, 2 50 

Essentials of Volumetric Analysis nmo, 1 25 

Spencer's Handbook for Chemists of Beet-sugar Houses i6mo, mcrocco, 3 00 

Handbook for Sugar Manufacturers and their Chemists. . i6mo, morocco, 2 00 

Stockbridge's Rocks and Soils 8vo, 2 50 

* Tillman's Elementary Lessons in Heat 8vo, 1 50 

* Descriptive General Chemistry 8vo, 3 00 

Treadwell's Qualitative Analysis. (Hall.) 8vo, 3 00 

Quantitative Analysis. (Hall.) 8vo, 4 00 

Turneaure and Russell's Public Water-supplies 8vo, 5 00 

Van Deventer's Physical Chemistry for Beginners. (Boltwood.) nmo, 1 50 

* Walke's Lectures on Explosives 8 Tr o, 4 00 

Washington's Manual of the Chemical Analysis of Rocks 8~o, 2 00 

Wassermann's Immune Sera : Haemolysins, Cytotoxins, and Precipitins. (Bol- 
duan.) nmo, 1 00 

Well's Laboratory Guide in Qualitative Chemical Analysis 8vo, 1 50 

Short Course in Inorganic Qualitative Chemical Analysis for Engineering 

Students nmo, 1 50 

Text-book of Chemical Arithmetic nmo, 1 25 

Whipple's Microscopy of Drinking-water 8vo, 3. 50 

Wilson's Cyanide Processes nmo, 1 50 

Chlorination Process nmo, 1 50 

Wulling's Elementary Course in Inorganic, Pharmaceutical, and Medical 

Chemistry nmo, 2 00 

CIVIL ENGINEERING. 
BRIDGES AND ROOFS. HYDRAULICS. MATERIALS OF ENGINEERING. 
RAILWAY ENGINEERING. 

Baker's Engineers' Surveying Instruments " nmo, 3 00 

Bixby's Graphical Computing Table Paper 19^X24! inches. 25 

** Burr's Ancient and Modern Engineering and the Isthmian Canal. (Postage, 

27 cents additional.) 8vo, 3 50 

Comstock's Field Astronomy for Engineers 8vo, 2 50 

Davis's Elevation and Stadia Tables 8vo, 1 00 

Elliott's Engineering for Land Drainage nmo, 1 50 

Practical Farm Drainage nmo, 1 00 

*Fiebeger's Treatise on Civil Engineering 8vo, 5 00 

Folwell's Sewerage. (Designing and Maintenance.) 8vo, 3 00 

Freitag's Architectural Engineering. 2d Edition, Rewritten 8vo, 3 50 

French and Ives's Stereotomy 8vo, 2 50 

Goodhue's Municipal Improvements nmo, 1 75 

Goodrich's Economic Disposal of Towns' Refuse 8vo, 3 50 

Gore's Elements of Geodesy 8vo, 2 50 

Hayford's Text-book of Geodetic Astronomy 8vo, 3 00 

Hering's Ready Reference Tables (Conversion Factors) i6mo, morocco, 2 50 

5 



Howe's Retaining Walls for Earth i2mo, i 25 

Johnson's (J. B.) Theory and Practice of Surveying Small 8vo, 4 00 

Johnson's (L. J.) Statics by Algebraic and Graphic Methods 8vo, 2 00 

Laplace's Philosophical Essay on Probabilities. (Truscott and Emory.) . i2mo, 2 00 

Mahan's Treatise on Civil Engineering. (1873.) (Wood.) 8vo, 5 00 

* Descriptive Geometry 8vo, 1 50 

Merriman's Elements of Precise Surveying and Geodesy 8vo, 2 50 

Elements of Sanitary Engineering 8vo, 2 00 

Merriman and Brooks's Handbook for Surveyors i6mo, morocco, 2 00 

Uugent's Plane Surveying 8vo, 3 50 

Ogden's Sewer Design i2mo, 2 00 

Patton's Treatise on Civil Engineering 8vo half leather, 7 50 

Reed's Topographical Drawing and Sketching 4to, 5 00 

Rideal's Sewage and the Bacterial Purification of Sewage 8vo, 3 50 

Siebert and Biggin's Modern Stone-cutting and Masonry 8yo, 1 50 

Smith's Manual of Topographical Drawing. (McMillan.) 8vo, 2 50 

Sondericker's Graphic Statics, with Applications to Trusses, Beams, and Arches. 

8vo, 2 00 

Taylor and Thompson's Treatise on Concrete, Plain and Reinforced 8vo, 5 00 

* Trautwine's Civil Engineer's Pocket-book i6mo, morocco, 5 00 

Wait's Engineering and Architectural Jurisprudence 8vo, 6 00 

Sheep, 6 50 
Law of Operations Preliminary to Construction in Engineering and Archi- 
tecture 8vo, 5 00 

Sheep, 5 50 

Law of Contracts 8vo, 3 00 

Warren's Stereotomy — Problems in Stone-cutting 8vo, 2 50 

Webb's Problems in the Use and Adjustment of Engineering Instruments. 

i6mo, morocco, 1 25 

* Wheeler s Elementary Course of Civil Engineering 8vo, 4 00 

Wilson's Topographic Surveying 8vo, 3 50 

BRIDGES AND ROOFS. 

Boiler's Practical Treatise on the Construction of Iron Highway Bridges. .8vo, 2 00 

* Thames River Bridge 4to, paper, 5 00 

Burr's Course on the Stresses in Bridges and Roof Trusses, Arched Ribs, and 

Suspension Bridges 8vo, 3 50 

Burr and Falk's Influence Lines for Bridge and Roof Computations. . . 8vo, 3 00 

Du Bois's Mechanics of Engineering. Vol. II Small 4to, 10 00 

Foster's Treatise on Wooden Trestle Bridges 4to, 5 00 

Fowler's Ordinary Foundations 8vo, 3 50 

Greene's Roof Trusses 8vo r 1 23 

Bridge Trusses 8vo, 2 50 

Arches in Wood, Iron, and Stone 8vo, 2 50 

Howe's Treatise on Arches 8vo, 4 00 

Design of Simple Roof-trusses in Wood and Steel 8vo, 2 00 

Johnson, Bryan, and Turneaure's Theory and Practice in the Designing of 

Modern Framed Structures Small 4to, 10 00 

Merriman and Jacoby's Text-book on Roofs and Bridges: 

Part I. Stresses in Simple Trusses 8vo, 

Part II. Graphic Statics 8vo, 

Part III. Bridge Design 8vo, 

Part TV. Higher Structures 8vo, 

Morison's Memphis Bridge 4to, 

Waddell's De Pontibus, a Pocket-book for Bridge Engineers. . i6mo, morocco, 

Specifications for Steel Bridges nmo, 

Wood's Treatise on the Theory of the Construction of Bridges and Roofs. .8vo, 
Wright's Designing of Draw-spans : 

Part I. Plate-girder Draws 8vo, 

Part II. Riveted-truss and Pin-connected Long-span Draws 8vo, 

Two parts in one volume 8vo, 

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HYDRAULICS. 

Bazin's Experiments upon the Contraction of the Liquid Vein Issuing from 

an Orifice. (Trautwine.) 8vo, 

Bovey's Treatise on Hydraulics 8vo, 

Church's Mechanics of Engineering 8vo, 

Diagrams of Mean Velocity of Water in Open Channels paper, 

Coffin's Graphical Solution of Hydraulic Problems i6mo, morocco, 

Flather's Dynamometers, and the Measurement of Povrer nmo, 

Folwell's Water-supply Engineering 8vo, 

Frizell's Water-power 8vo, 

Fuertes's Water and Public Health nmo, 

Water-filtration Works nmo, 

Ganguillet and Kutter's General Formula for the Uniform Flow of Water in 

Rivers and Other Channels. (Hering and Trautwine.) 8vo, 

Hazen's Filtration of Public Water-supply 8vo, 

Hazlehurst's Towers and Tanks for Water-works 8vo, 

Herschel's 115 Experiments on the Carrying Capacity of Large, Riveted, Metal 

Conduits 8vo, 

Mason's Water-supply. (Considered Principally from a Sanitary Standpoint.) 

8vo, 

Merriman's Treatise on Hydraulics 8vo, 

* Michie's Elements of Analytical Mechanics 8vo, 

Schuyler's Reservoirs for Irrigation, Water-power, and Domestic Water- 
supply Large 8vo , 

** Thomas and Watt's Improvement of Rivers. (Post., 44c. additional.). 4to, 

Turneaure and Russell's Public Water-supplies 8vo, 

Wegmann's Design and Construction of Dams 4to, 

Water-supply of the City of New York from 1658 to 1895 4to, 

Williams and Hazen's Hydraulic Tables 8vo, 

Wilson's Irrigation Engineering Small 8vo, 

Wolff's Windmill as a Prime Mover 8vo, 

Wood's Turbines. 8vo, 

Elements of Analytical Mechanics 8vo, 

MATERIALS OF ENGINEERING. 

Baker's Treatise on Masonry Construction 8vo, 

Roads and Pavements . .8vo, 

Black's United States Public Works Oblong 4:0, 

Bovey's Strength of Materials and Theory of Structures 8vo, 

Burr's Elasticity and Resistance of the Materials of Engineering 8vo, 

Byrne's Highway Construction 8vo, 

Inspection of the Materials and Workmanship Employed in Construction. 

i&mo, 

Church's Mechanics of Engineering 8vo, 

Du Bois's Mechanics of Engineering. Vol. I Small 4to, 

^Eckel's Cements, Limes, and Plasters 8vo, 

Johnson's Materials of Construction Large 8vo, 

Fowler's Ordinary Foundations , 8vo, 

Keep's Cast Iron 8vo, 

Lanza's Applied Mechanics 8vo, 

Marten's Handbook on Testing Materials. (Henning.) 2 vols 8vo, 

Merrill's Stones for Building and Decoration 8vo, 

Merriman's Mechanics of Materials. 8vo, 

Strength of Materials nmo, 

Metcalf's Steel. A Manual for Steel-users nmo, 

Patton's Practical Treatise on Foundations 8vo, 

Richardson's Modern Asphalt Pavements 8vo, 

Richey's Handbook for Superintendents of Construction i6mo, mor., 

Rockwell's Roads and Pavements in France i2mo, 

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Sabin's Industrial and Artistic Technology of Paints and Varnish 8vo, 

Smith's Materials of Machines nmo, 

Snow's Principal Species of Wood 8vo, 

Spalding's Hydraulic Cement i2mo, 

Text-book on Roads and Pavements nmo, 

Taylor and Thompson's Treatise on Concrete, Plain and Reinforced 8vo, 

Thurston's Materials of Engineering. 3 Parts 8vo, 

Part I. Non-metallic Materials of Engineering. and Metallurgy 8vo, 

Part II. Iron and Steel 8vo, 

Part III. A Treatise on Brasses, Bronzes, and Other Alloys and their 

Constituents 8vo, 

Thurston's Text-book of the Materials of Construction 8vo, 

Tillson's Street Pavements and Paving Materials 8vo, 

Waddell's De Pontibus. ( A Pocket-book for Bridge Engineers.) . . i6mo, mor., 

Specifications for Stc i Bridges. . i2mo, 

Wood's (De V.) Treatise on the Resistance of Materials, and an Appendix on 

the Preservation of Timber 8vo, 

Wood's (De V.) Elements of Analytical Mechanics 8vo, 

Wood's (M. P.) Rustless Coatings: Corrosion and Electrolysis of Iron and 

Steel . 8vo, 4 00 

RAILWAY ENGINEERING. 

Andrew's Handbook for Street Railway Engineers 3x5 inches, morocco, 

Berg's Buildings and Structures of American Railroads 4to, 

Brook's Handbook of Street Railroad Location i6mo, morocco, 

Butt's Civil Engineer's Field-book i6mo, morocco, 

Crandall's Transition Curve i6mo, morocco, 

Railway and Other Earthwork Tables 8vo, 

Dawson's "Engineering" and Electric Traction Pocket-book. . i6mo, morocco, 
Dredge's History of the Pennsylvania Railroad: (1879) Paper, 

* Drinker's Tunnelling, Explosive Compounds, and Rock Drills. 4to, half mor., 

Fisher's Table of Cubic Yards Cardboard, 

Godwin's Railroad Engineers' Field-book and Explorers' Guide. . . i6mo, mor., 

Howard's Transition Curve Field-book i6mo, morocco, 

Hudson's Tables for Calculating the Cubic Contents of Excavations and Em- 
bankments 8vo, 

Molitor and Beard's Manual for Resident Engineers i6mo, 

Nagle's Field Manual for Railroad Engineers i6mo, morocco, 

Philbrick's Field Manual for Engineers i6mo, morocco, 

Searles's Field Engineering i6mo, morocco, 

Railroad Spiral i6mo, morocco, 

Taylor's Prismoidal Formulae and Earthwork 8vo, 

* Trautwine's Method of Calculating the Cube Contents of Excavations and 

Embankments by the Aid of Diagrams 8vo, 

The Field Practice of Laying Out Circular Curves for Railroads. 

s i2mo, morocco, 

Cross-section Sheet , Paper, 

Webb's Railroad Construction i6mo, morocco, 

Wellington's Economic Theory of the Location of Railways Small 8vo, 

DRAWING. 

Barr's Kinematics of Machinery 8vo, 

* Bartlett's Mechanical Drawing 8vo, 

* " " " Abridged Ed 8vo, 

Coolidge's Manual of Drawing 8vo, paper 

Coolidge and Freeman's Elements of General Drafting for Mechanical Engi- 
neers Oblong 4to, 

Durley's Kinematics of Machines 8vo, 

Emch's Introduction to Projective Geometry and its Applications 8vo. 

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Hill's Text-book on Shades and Shadows, and Perspective 8vo, 

Jamison's Elements of Mechanical Drawing 8vo, 

Advanced Mechanical Drawing 8vo, 

Jones's Machine Design: 

Part I. Kinematics of Machinery 8vo, 

Part II. Form, Strength, and Proportions of Parts. . . ., 8vo, 

MacCord's Elements of Descriptive Geometry 8vo, 

Kinematics ; or, Practical Mechanism 8vo, 

Mechanical Drawing 4to, 

Velocity Diagrams 8vo, 

* Mahan's Descriptive Geometry and Stone-cutting 8vo, 

Industrial Drawing. (Thompson.) 8vo, 

Moyer's Descriptive Geometry 8vo, 

Reed's Topographical Drawing and Sketching 4to, 

Reid's Course in Mechanical Drawing 8vo, 

Text-book of Mechanical Drawing and Elementary Machine Design. 8vo, 

Robinson's Principles of Mechanism 8vo, 

Schwamb and Merrill's Elements of Mechanism 8vo, 

Smith's Manual of Topographical Drawing. (McMillan.) 8vo, 

Warren's Elements of Plane and Solid Free-hand Geometrical Drawing, nmo, 

Drafting Instruments and Operations nmo, 

Manual of Elementary Projection Drawing . . i2mo, 

Manual of Elementary Problems in the Linear Perspective of Form and 

Shadow i2mo, 

Plane Problems in Elementary Geometry nmo, 

Primary Geometry nmo, 

Elements of Descriptive Geometry, Shadows, and Perspective 8vo, 

General Problems of Shades and Shadows 8vo, 

Elements of Machine Construction and Drawing 8vo, 

Problems, Theorems, and Examples in Descriptive Geometry 8vo, 

Weisbach's Kinematics and Power of Transmission. (Hermann and Klein)8vo , 

Whelpley's Practical Instruction in the Ail of Letter Engraving nmo, 

Wilson's (H. M.) Topographic Surveying 8vo, 

Wilson's (V. T.) Free-hand Perspective 8vo, 

Wilson's (V. T.) Free-hand Lettering 8vo, 

Woolf's Elementary Course in Descriptive Geometry Large 8vo, 



ELECTRICITY AND PHYSICS. 

Anthony and Brackett's Text-book of Physics. (Magie.) Small 8vo, 

Anthony's Lecture-notes on the Theory of Electrical Measurements. . . . nmo, 
Benjamin's History of Electricity, 8vo, 

Voltaic Cell 8vo, 

Classen's Quantitative Chemical Analysis by Electrolysis. (Boltwood.).Svo, 

Crehore and Squier's Polarizing Photo-chronograph 8vo, 

Dawson's "Engineering" and Electric Traction Pocket-book. i6mo, morocco, 
Dolezalek's Theory of the Lead Accumulator (Storage Battery). (Von 

Ende.) nmo, 

Duhem's Thermodynamics and Chemistry. (Burgess.) 8vo, 

Flather's Dynamometers, and the Measurement of Power nmo, 

Gilbert's De Magnete. (Mottelay.) 8vo, 

Hanchett's Alternating Currents Explained nmo, 

Hering's Ready Reference Tables (Conversion Factors) i6mo, morocco, 

Holman's Precision of Measurements 8vo, 

Telescopic Mirror-scale Method, Adjustments, and Tests. . . .Large 8vo, 

Kinzbrunner's Testing of Continuous-Current Machines 8vo, 

Landauer's Spectrum Analysis. (Tingle.) 8vo, 

Le Chatelien's High-temperature Measurements. (Boudouard — Burgess.) nmo. 
Lob's Electrolysis and Electrosynthesis of Organic Compounds. (Lorenz.) nmo, 

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* Lyons's Treatise on Electromagnetic Phenomena. Vols. I. and II. 8vo, each, 6 00 

* Michie's Elements of Wave Motion Relating to Sound and Light 8vo, 4 00 

Niaudet's Elementary Treatise on Electric Batteries. (Fishback.) i2mo, 2 50 

* Rosenberg's Electrical Engineering. (Haldane Gee — Kinzbrunner.). . 8vo, 1 50 

Ryan, Norris, and Hoxie's Electrical Machinery. Vol. 1 8vo, 2 50 

Thurston's Stationary Steam-engines 8vo, 2 50 

* Tillman's Elementary Lessons in Heat 8vo, 1 50 

Tory and Pitcher's Manual of Laboratory Physics Small 8vo, 2 00 

Ulke's Modern Electrolytic Copper Refining 8vo, 3 00 

LAW. 

* Davis's Elements of Law 8vo, 

* Treatise on the Military Law of United States 8vo, 

* Sheep, 

Manual for Courts-martial i6mo, morocco, 

Wait's Engineering and Architectural Jurisprudence 8vo, 

Sheep, 
Law of Operations Preliminary to Construction in Engineering and Archi- 
tecture 8vo, 

Sheep, 

Law of Contracts 8vo, 

Winthrop's Abridgment of Military Law i2mo, 

MANUFACTURES. 

Bernadou's Smokeless Powder — Nitro-cellulose and Theory of the Cellulose 

Molecule nmo, 

Bolland's Iron Founder nmo, 

"The Iron Founder," Supplement nmo, 

Encyclopedia of Founding and Dictionary of Foundry Terms Used in the 

Practice of Moulding ' nmo, 

Eissler's Modern High Explosives 8vo, 

Effront's Enzymes and their Applications. (Prescott.) 8vo, 

Fitzgerald's Boston Machinist nmo, 

Ford's Boiler Making for Boiler Makers i8mo, 

Hopkin's Oil-chemists' Handbook 8vo, 

Keep's Cast Iron 8vo, 

Leach's The Inspection and Analysis of Food with Special Reference to State 

Control Large 8vo, 

Matthews's The Textile Fibres 8vo, 

Metcalf's Steel. A Manual for Steel-users nmo, 

Metcalfe's Cost of Manufactures — And the Administration of Workshops. 8vo, 

Meyer's Modern Locomotive Construction 4to, 

Morse's Calculations used in Cane-sugar Factories i6mo, morocco, 

* Reisig's Guide to Piece-dyeing 8vo, 

Sabin's Industrial and Artistic Technology of Paints and Varnish 8vo, 

Smith's Press-working of Metals 8vo, 

Spalding's Hydraulic Cement nmo, 

Spencer's Handbook for Chemists of Beet-sugar Houses. . . . i6mo, morocco, 

Handbook for Sugar Manufacturers and their Chemists. . i6mo, morocco, 

Taylor and Thompson's Treatise on Concrete, Plain and Reinforced 8vo, 

Thurston's Manual of Steam-boilers, their Designs, Construction and Opera- 
tion 8vo, 

* Walke's Lectures on Explosives 8vo, 

Ware's Manufacture of Sugar. (In press.) 

West's American Foundry Practice nmo, 

Moulder's Text-book nmo, 

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Wolff's Windmill as a Prime Mover : 8vo, 3 00 

Wood's Rustless Coatings: Corrosion and Electrolysis of Iron and Steel. .8vo, 4 00 



MATHEMATICS. 

Baker's Elliptic Functions 8vo, 

* Bass's Elements of Differential Calculus nmo, 

Briggs's Elements of Plane Analytic Geometry nmo, 

Compton's Manual of Logarithmic Computations nmo, 

Davis's Introduction to the Logic of Algebra 8vo, 

* Dickson's College Algebra Large nmo, 

* Introduction to the Theory of Algebraic Equations Large nmo, 

Emch's Introduction to Projective Geometry and its Applications 8vo, 

Halsted's Elements of Geometry 8vo, 

Elementary Synthetic Geometry 8vo, 

Rational Geometry nmo, 

♦Johnson's (J. B.) Three-place Logarithmic Tables: Vest-pocket size, paper, 

100 copies for 

* Mounted on heavy cardboard, 8X10 inches, 

10 copies for 
Johnson's (W. W.) Elementary Treatise on Differential Calculus. .Smah 8vo, 
Johnson's (W. W.) Elementary Treatise on the Integral Calculus. Small 8 vo, 

Johnson's (W. W.) Curve Tracing in Cartesian Co-ordinates nmo, 

Johnson's (W. W.) Treatise on Ordinary and Partial Differential Equations. 

Small 8vo, 
Johnson's (W. W.) Theory of Errors and the Method of Least Squares. nmo, 

* Johnson's (W. W.) Theoretical Mechanics nmo, 

Laplace's Philosophical Essay on Probabilities. (Truscott and Emory.) . nmo, 

* Ludlow and Bass. Elements of Trigonometry and Logarithmic and Other 

Tables 8vo, 

Trigonometry and Tables published separately Each, 

* Ludlow's Logarithmic and Trigonometric Tables 8vo, 

Maurer's Technical Mechanics 8vo, 

Merriman and Woodward's Higher Mathematics 8vo, 

Merriman's Method of Least Squares 8vo, 

Rice and Johnson's Elementary Treatise on the Differential Calculus. . Sm. 8vo, 

Differential and Integral Calculus. 2 vols, in one Small 8vo, 

Wood's Elements of Co-ordinate Geometry. 8vo, 

Trigonometry: Analytical, Plane, and Spherical nmo, 



MECHANICAL ENGINEERING. 

MATERIALS OF ENGINEERING, STEAM-ENGINES AND BOILERS. 

Bacon's Forge Practice nmo, 1 50 

Baldwin's Steam Heating for Buildings nmo, 2 50 

Barr's Kinematics of Machinery 8vo, 2 50 

* Bartlett's Mechanical Drawing 8vo, 3 00 

* " " " Abridged Ed 8vo, 1 50 

Benjamin's Wrinkles and Recipes nmo, 2 00 

Carpenter's Experimental Engineering 8vo, 6 00 

Heating and Ventilating Buildings 8vo, 4 00 

Cary's Smoke Suppression in Plants using Bituminous Coal. (In Prepara- 
tion.) 

Clerk's Gas and Oil Engine Small 8vo, 4 00 

Coolidge's Manual of Drawing 8vo, paper, 1 00 

Coolidge and Freeman's Elements of General Drafting for Mechanical En- 
gineers Oblong 4to, 2 50 

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Cromwell's Treatise on Toothed Gearing i2mo, 

Treatise on Belts and Pulleys .' i2mo, 

Durley's Kinematics of Machines 8vo, 

Flather's Dynamometers and the Measurement of Power. i2mo, 

Rope Driving i2mo, 

Gill's Gas and Fuel Analysis for Engineers i2mo, 

Hall's Car Lubrication i2mo, 

Hering's Ready Reference Tables (Conversion Factors) i6mo, morocco, 

Hutton's The Gas Engine 8vo, 

Jamison's Mechanical Drawing 8vo, 

Jones's Machine Design : 

Part I. Kinematics of Machinery 8vo, 

Part II. Form, Strength, and Proportions of Parts 8vo, 

Kent's Mechanical Engineers' Pocket-book i6mo, morocco, 

Kerr's Power and Power Transmission. 8vo, 

Leonard's Machine Shop, Tools, and Methods 8vo, 

*Lorenz's Modern Refrigerating Machinery. (Pope, Haven, and Dean.) . . 8vo, 

MacCord's Kinematics; or, Practical Mechanism 8vo, 

Mechanical Drawing 4to, 

Velocity Diagrams 8vo, 

Mahan's Industrial Drawing. (Thompson.) 8vo, 

Poole s Calorific Power of Fuels 8vo, 

Reid's Course in Mechanical Drawing 8vo, 

Text-book of Mechanical Drawing and Elementary Machine Design. 8vo, 

Richard's Compressed Air i2mo, 

Robinson's Principles of Mechanism 8vo, 

Schwamb and Merrill's Elements of Mechanism 8vo, 

Smith's Press-working of Metals 8vo, 

Thurston's Treatise on Friction and Lost Work in Machinery and Mill 

Work 8vo, 

Animal as a Machine and Prime Motor, and the Laws of Energetics, nmo, 

Warren's Elements of Machine Construction and Drawing .8vo, 

Weisbach's Kinematics and the Power of Transmission. (Herrmann — 

Klein.) . 8vo, 

Machinery of Transmission and Governors. (Herrmann — Klein.). .8vo, 

Wolff's Windmill as a Prime Mover 8vo, 

Wood's Turbines 8vo, 



MATERIALS OF ENGINEERING. 

Bovey's Strength of Materials and Theory of Structures 8vo, 7 50 

Burr's Elasticity and Resistance of the Materials of Engineering. 6th Edition. 

Reset 8vo, 

Church's Mechanics of Engineering , .8vo, 

Johnson's Materials of Construction 8vo, 

Keep's Cast Iron 8vo, 

Lanza's Applied Mechanics 8vo, 

Martens's Handbook on Testing Materials. (Henning.) 8vo, 

Merriman's Mechanics of Materials. 8vo, 

Strength of Materials nmo, 

Metcalf's Steel. A manual for Steel-users i2mo. 

Sabin's Industrial and Artistic Technology of Paints and Varnish 8vo, 

Smith's Materials of Machines i2mo, 

Thurston's Materials of Engineering 3 vols., 8vo, 

Part II. Iron and Steel 8vo, 

Part IH. A Treatise on Brasses, Bronzes, and Other Alloys and their 
Constituents 8vo, 

Text-book of the Materials of Construction 8vo, 

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Wood's (De V.) Treatise on the Resistance of Materials and an Appendix on 

the Preservation of Timber 8vo, 2 oO 

Wood's (De V.) Elements of Analytical Mechanics 8vo, 3 00 

Wood's (M. P.) Rustless Coatings: Corrosion and Electrolysis of Iron and 

Ste»L 8vo, 4 00 



STEAM-ENGINES AND BOILERS. 



Berry's Temperature-entropy Diagram i2mo, 

Carnot's Reflections on the Motive Power of Heat. (Thurston.) nmo, 

Dawson's "Engineering" and Electric Traction Pocket-book. . . .i6mo, mor., 

Ford's Boiler Making for Boiler Makers i8rno, 

Goss's Locomotive Sparks 8vo, 

Hemenway's Indicator Practice and Steam-engine Economy i2mo, 

Hutton's Mechanical Engineering of Power Plants 8vo, 

Heat and Heat-engines 8vo, 

Kent's Steam boiler Economy 8vo, 

Kneass's Practice and Theory of the Injector 8vo, 

MacCord's Slide-valves 8vo, 

Meyer's Modern Locomotive Construction 4to, 

Peabody's Manual of the Steam-engine Indicator nmo. 

Tables of the Properties of Saturated Steam and Other Vapors 8vo, 

Thermodynamics of the Steam-engine and Other Heat-engines 8vo, 

Valve-gears for Steam-engines 8vo, 

Peabody and Miller's Steam-boilers 8vo, 

Pray's Twenty Years with the Indicator Large 8vo, 

Pupin's Thermodynamics of Reversible Cycles in Gases and Saturated Vapors. 

(Osterberg.) nmo, 

Reagan's Locomotives: Simple Compound, and Electric nmo, 

Rontgen's Principles of Thermodynamics. (Du Bois.) 8vo, 

Sinclair's Locomotive Engine Running and Management nmo, 

Smart's Handbook of Engineering Laboratory Practice nmo, 

Snow's Steam-boiler Practice 8vo, 

Spangler's Valve-gears 8vo, 

Notes on Thermodynamics nmo, 

Spangler, Greene, and Marshall's Elements of Steam-engineering 8vo, 

Thurston's Handy Tables 8vo, 

Manual of the Steam-engine 2 vols., 8vo, 

Part I. History, Structure, and Theory 8vo, 

Part H. Design, Construction, and Operation. . 8vo, 

Handbook of Engine and Boiler Trials, and the Use of the Indicator and 

the Prony Brake 8vo, 

Stationary Steam-engines ' 8vo, 

Steam-boiler Explosions in Theory and in Practice .nmo, 

Manual of Steam-boilers, their Designs, Construction, and Operation 8vo, 

Weisbach's Heat, Steam, and Steam-engines. (Du Bois.) 8vo, 

Whitham's Steam-engine Design 8vo, 

Wilson's Treatise on Steam-boilers. (Flather.) i6mo, 

Wood's Thermodynamics, Heat Motors, and Refrigerating Machines. . .8vo, 



MECHANICS AND MACHINERY. 

Barr's Kinematics of Machinery 8vo, 2 50 

Bovey's Strength of Materials and Theory of Structures 8vo, 7 50 

Chase's The Art of Pattern-making „ nmo, 2 50 

Churches Mechanics of Engineering 8vo, 6 00 

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Church's Notes and Examples in Mechanics 8vo, 2 00 

Compton's First Lessons in Metal- working i2mo, 1 50 

Compton and De Groodt's The Speed Lathe i2mo, 1 50 

Cromwell's Treatise on Toothed Gearing; i2mo, 1 50 

Treatise on Belts and Pulleys i2mo, 1 50 

Dana's Text-book of Elementary Mechanics for Colleges and Schools, .nmo, 1 50 

Dingey's Machinery Pattern Making i2mo, 2 00 

Dredge's Record of the Transportation Exhibits Building of the World's 

Columbian Exposition of 1893 4to half mcrocco, 5 00 

Du Bois's Elementary Principles of Mechanics: 

Vol. I. Kinematics 8vo, 

Vol. II. Statics ; 8vo, 

Vol. III. Kinetics 8vo, 

Mechanics of Engineering. Vol. I Small 4to, 

Vol. II Small 4to, 

Durley's Kinematics of Machines 8vo, 

Fitzgerald's Boston Machinist i6mo, 

Flather's Dynamometers, and the Measurement of Power i2mo, 

Rope Driving i2mo, 

Goss's Locomotive Sparks 8vo, 

Hall's Car Lubrication i2mo, 

Holly's Art of Saw Filing i8mo, 

James's Kinematics of a Point and the Rational Mechanics of a Particle. Sm.8vo,2 00 

* Johnson's (W. W.) Theoretical Mechanics i2mo, 

Johnson's (L. J.) Statics by Graphic and Algebraic Methods 8vo, 

Jones's Machine Design : 

Part I. Kinematics of Machinery 8vo, 

Part II. Form, Strength, and Proportions of Parts 8vo, 

Kerr's Power and Power Transmission 8vo, 

Lanza's Applied Mechanics 8vo, 

Leonard's Machine Shop, Tools, and Methods 8vo, 

*Lorenz's Modern Refrigerating Machinery. (Pope, Haven, and Dcan.).8vo, 

MacCord's Kinematics; or, Practical Mechanism 8vo, 

Velocity Diagrams 8vo, 

Maurer's Technical Mechanics 8vo, 

Merriman's Mechanics of Materials 8vo, 

* Elements of Mechanics i2mo, 

* Michie's Elements of Analytical Mechanics 8vo, 

Reagan's Locomotives: Simple, Compound, and Electric i2mo > 

Reid's Course in Mechanical Drawing 8vo, 

Text-book of Mechanical Drawing and Elementary Machine Design. 8vo, 

Richards's Compressed Air i2mo, 

Robinson's Principles of Mechanism 8vo, 

Ryan, Norris, and Hoxie's Electrical Machinery. Vol. 1 8vo, 

Schwamb and Merrill's Elements of Mechanism .8vo, 

Sinclair's Locomotive-engine Running and Management i2mo, 

Smith's (O.) Press-working of Metals 8vo, 

Smith's (A. W.) Materials of Machines i2mo, 

Spangler, Greene, and Marshall's Elements of Steam-engineering 8vo, 

Thurston's Treatise on Friction and Lost Work in Machinery and Mill 
Work 8vo, 

Animal as a Machine and Prime Motor, and the Laws of Energetics. 

i2mo, 

Warren's Elements of Machine Construction and Drawing 8vo, 

Weisbach's Kinematics and Power of Transmission. ( Herrmann — Klein. ) . 8vo , 

Machinery of Transmission and Governors. (Herrmann — Klein. ).8vo, 
Wood's Elements of Analytical Mechanics 8vo, 

Principles of Elementary Mechanics i2mo, 

Turbines 8vo v 

The World's Columbian Eaposition of 1893 4to, 

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METALLURGY. 

Egleston's Metallurgy of Silver, Gold, and Mercury: 

Vol. t Silver 8vo y 

Vol. II. Gold and Mercury 8vo, 

** Iles's Lead-smelting. (Postage 9 cents additional.) nmo, 

Keep's Cast Iron 8vo, 

Kunhardt's Practice of Ore Dressing in Europe 8vo, 

Le Chatelier's High-temperature Measurements. (Boudouard — Burgess. )i2mo, 

Metcalf's Steel. A Manual for Steel-users ' nmo, 

Smith's Materials of Machines i2mo, 

Thurston's Materials of Engineering. In Three Parts 8vo. 

Part II. Iron and Steel 8vo, 

Part III. A Treatise on Brasses, Bronzes, and Other Alloys and their 

Constituents 8vo, 

Ulke's Modern Electrolytic Copper Refining. 8vo, 

MINERALOGY. 

Barringer's Description of Minerals of Commercial Value. Oblong, morocco, 

Boyd's Resources of Southwest Virginia 8vo, 

Map of Southwest Virignia Pocket-book form. 

Brush's Manual of Determinative Mineralogy. (Penfield.) 8vo, 

Chester's Catalogue of Minerals 8vo, paper, 

Cloth, 

Dictionary of the Names of Minerals 8vo, 

Dana's System of Mineralogy Large 8vo, half leather, 12 50 

First Appendix to Dana's New " System of Mineralogy." Large 8vo, 

Text-book, of Mineralogy. 8vo, 

Minerals and How to Study Them i2mo, 

Catalogue of American Localities of Minerals. '. Large 8vo, 

Manual of Mineralogy and Petrography i2mo, 

Douglas's Untechnical Addresses on Technical Subjects nmo, 

Eakle's Mineral Tables 8vo, 

Egleston's Catalogue of Minerals and Synonyms 8vo, 

Hussak's The Determination of Rock-forming Minerals. (Smith.) .Small 8vo, 
Merrill's Non-metallic Minerals: Their Occurrence and Uses 8vo, 

* Penfield's Notes on Determinative Mineralogy and Record of Mineral Tests. 

8vo paper, o 50 
Rosenbusch's Microscopical Physiography of the Rock-making Minerals. 
(Iddings.) , 8vo. 

* Tillman's Text-book of Important Minerals and Rocks , .8vo. 

Williams's Manual of Lithology 8vo, 

MINING. 

Beard's Ventilation of Mines i2mo, 

Boyd's Resources of Southwest Virginia 8vo. 

Map of Southwest Virginia Pocket book form, 

Douglas's Untechnical Addresses on Technical Subjects i2mo. 

* Drinker's Tunneling, Explosive Compounds, and Rock Drills. 4to,hf. mor . 

Eissler's Modern High Explosives 8vo. 

Fowler's Sewage Works Analyses , . 12 mo, 

Goodyear's Coal-mines of the Western Coast of the United States i2mo, 

Ihlseng's Manual of Mining 8vo* 

** Iles's Lead-smelting. (Postage 9c. additional.). i2mo. 

Kunhardt's Practice of Ore Dressing in Europe 8vo„ 

O'Driscoll's Notes on the Treatment of Gold Ores .8vo. 

* Walke's Lectures on Explosives „ , 8vo, 

Wilson's Cyanide Processes , I2m0j 

Chlorination Process. i2mo, 

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Wilson's Hydraulic and Placer Mining i2mo 

Treatise on Practical and Theoretical Mine Ventilation i2mo', 

SANITARY SCIENCE. 

Bashore's Sanitation of a Country House I2mo 

Folwell's Sewerage. (Designing, Construction, and Maintenance.) 8vo, 

Water-supply Engineering 8vo, 

Fuertes's Water and Public Health i2mo, 

Water-filtration Works i2mo, 

Gerhard's Guide to Sanitary House-inspection i6mo, 

Goodrich's Economic Disposal of Town's Refuse Demy 8vo, 

Hazen's Filtration of Public Water-supplies 8vo, 

Leach's The Inspection and Analysis of Food with Special Reference to State 

Control 8vo, 

Mason's Water-supply. (Considered principally from a Sanitary Standpoint) 8vo, 

Examination of Water. (Chemical and Bacteriological.) i2mo, 

Merriman's Elements of Sanitary Engineering 8vo, 

Ogden's Sewer Design i2mo, 

Prescott and Winslow's Elements of Water Bacteriology, with Special Refer- 
ence to Sanitary Water Analysis i2mo, 

* Price's Handbook on Sanitation i2mo, 

Richards's Cost of Food. A Study in Dietaries i2mo, 

Cost of Living as Modified by Sanitaiy Science i2mo, 

Richards and Woodman's Air, Water, and Food from a Sanitary Stand- 
point 8vo, 

* Richards and Williams's The Dietary Computer 8vo, 

Rideal's Sewage and Bacterial Purification of Sewage 8vo, 

Turneaure and Russell's Public Water-supplies : . . .8vo, 

Von Behring's Suppressipn of Tuberculosis. (Bolduan.) . . .- i2mo, 

Whipple's Microscopy of Drinking-water 8vo, 

Woodhull's Notes on Military Hygiene i6mo, i 50 

MISCELLANEOUS. 

De Fursac's Manual of Psychiatry. (Rosanoff and Collins.). .. .Large i2mo, 2 50 
Emmons's Geological Guide-book of the Rocky Mountain Excursion of the 

International Congress of Geologists Large 8vo, 

Ferrel's Popular Treatise on the Winds 8vo. 

Haines's American Railway Management nmo, 

Mott's Composition, Digestibility, and Nutritive Value of Food. Mounted chart, 

Fallacy of the Present Theory of Sound .••••' i6mo, 

Ricketts's History of Rensselaer Polytechnic Institute, 1824-1894. .Small 8vo, 

Rostoski's Serum Diagnosis. (Bolduan.) nmo, 

Rotherham's Emphasized New Testament Large 8vo, 

Steel's Treatise on the Diseases of the Dog .8vo, 

Totten's Important Question in Metrology 8vo, 

The World's Columbian Exposition of 1893 410, 

Von Behring's Suppression of Tuberculosis. (Bolduan.) nmo, 

Winslow's Elements of Applied Microscopy nmo, 

Worcester and Atkinson. Small Hospitals, Establishment and Maintenance; 

Suggestions for Hospital Architecture : Plans for Small Hospital. 1 2mo, 1 25 

HEBREW AND CHALDEE TEXT-BOOKS. 

Green's Elementary Hebrew Grammar ■ i2mo, 1 25 

Hebrew Chrestomathy 8vo, 2 00 

Gesenius's Hebrew and Chaldee Lexicon tr the Old Testament Scriptures. 

(Tregelles.) Small 4to, half morocco, 5 00 

Letteris's Hebrew Bible 8vo, 2 25 

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